Techniques of Integration

Study Sheet

Techniques of Integration

Everything you need for AP Calculus BC, Topic 7

uu-Substitution Review

Concept
The idea: reverse the chain rule

If you can write an integrand in the form f(g(x)) g′(x)f(g(x))\,g'(x), let u=g(x)u=g(x) so that du=g′(x) dxdu=g'(x)\,dx. The integral becomes ∫f(u) du\displaystyle\int f(u)\,du. Look for an “inside” function whose derivative also appears (up to a constant factor). For a definite integral, either change the limits to uu-values or convert back to xx before substituting the original limits.

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)

Substitution reverses the chain rule, integration by parts reverses the product rule, partial fractions split rational functions into pieces that integrate to logs and arctangents, and trigonometric substitutions remove square roots of quadratics. Improper integrals are limits of ordinary ones. The skill is recognizing which reversal fits.

Trickier substitutions to watch for: when only part of xx gets absorbed, you may need to solve for the leftover. For ∫xx−1 dx\displaystyle\int x\sqrt{x-1}\,dx, let u=x−1u=x-1, so x=u+1x=u+1 and dx=dudx=du.

Example
Solving for the leftover factor
∫xx−1 dx=∫(u+1)u du=∫(u3/2+u1/2) du=25u5/2+23u3/2+C=25(x−1)5/2+23(x−1)3/2+C.\begin{aligned} \int x\sqrt{x-1}\,dx &=\int (u+1)\sqrt{u}\,du=\int\big(u^{3/2}+u^{1/2}\big)\,du\\ &=\tfrac{2}{5}u^{5/2}+\tfrac{2}{3}u^{3/2}+C =\tfrac{2}{5}(x-1)^{5/2}+\tfrac{2}{3}(x-1)^{3/2}+C. \end{aligned}
Side note
Where it goes wrong

Changing variables in a definite integral without changing the limits, or back-substituting and then using the old limits. Pick one: convert the limits, or return to xx — not both.

Example
A substitution that clears a denominator
∫ln⁡xx dx,u=ln⁡x, du=1x dx  ⟹  ∫u du=12u2+C=12(ln⁡x)2+C.\int\frac{\ln x}{x}\,dx,\quad u=\ln x,\ du=\tfrac{1}{x}\,dx \;\Longrightarrow\;\int u\,du=\tfrac12 u^2+C=\tfrac12(\ln x)^2+C.
Tip

Tip: After substituting, every xx and dxdx must disappear. If an xx is left over, solve the uu-equation for xx and substitute it in — don't guess.

Try it
Quick check

∫2x (x2+1)5 dx\displaystyle\int 2x\,(x^2+1)^5\,dx

Work it on paper before revealing the answer.

Show the answer

Answer: (x2+1)66+C\dfrac{(x^2+1)^6}{6}+C

Integration by Parts

Concept
The formula and LIATE

Integration by parts reverses the product rule:

∫u dv=uv−∫v du.\int u\,dv = uv-\int v\,du.

Choose uu to be the factor that gets simpler when differentiated. The LIATE order tells you which factor to call uu (earlier = better choice for uu):

Logarithmic → Inverse trig → Algebraic → Trig → Exponential.\textbf{L}\text{ogarithmic}\ \to\ \textbf{I}\text{nverse trig}\ \to\ \textbf{A}\text{lgebraic}\ \to\ \textbf{T}\text{rig}\ \to\ \textbf{E}\text{xponential}.

Whatever is left (together with dxdx) becomes dvdv.

Reminder — Substitution and integration by parts:  ∫f(g(x))g′(x) dx=∫f(u) du,∫u dv=uv−∫v du\int f(g(x))g'(x)\,dx=\int f(u)\,du, \qquad \int u\,dv = uv-\int v\,du

Integrate both sides of the product rule (uv)′=u′v+uv′(uv)'=u'v+uv' and rearrange: ∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du. The formula trades one integral for another, so it helps only when ∫v du\int v\,du is simpler, which is what LIATE is for: choose as uu the factor that becomes simpler when differentiated.

Example
A full parts calculation

Evaluate ∫xex dx\displaystyle\int x e^{x}\,dx. By LIATE, Algebraic beats Exponential, so u=xu=x, dv=ex dxdv=e^{x}\,dx.

u=x⇒du=dx,dv=ex dx⇒v=ex.∫xex dx=xex−∫ex dx=xex−ex+C.\begin{aligned} u=x &\quad\Rightarrow\quad du=dx,\\ dv=e^{x}\,dx &\quad\Rightarrow\quad v=e^{x}.\\[2pt] \int x e^{x}\,dx &= xe^{x}-\int e^{x}\,dx = xe^{x}-e^{x}+C. \end{aligned}

Check: ddx(xex−ex)=ex+xex−ex=xex.\frac{d}{dx}\big(xe^x-e^x\big)=e^x+xe^x-e^x=xe^x. ✓

Repeated parts & the tabular method. When uu is a polynomial, you apply parts once per degree. The tabular method organizes this: differentiate uu down to 00 in one column, antidifferentiate dvdv in the other, and multiply along the diagonals with alternating signs +,−,+,…+,-,+,\dots

Example
Tabular method for ∫x2cos⁡x dx\displaystyle\int x^2\cos x\,dx
signD (u)I (dv)+x2cos⁡x−2xsin⁡x+2−cos⁡x−0−sin⁡x\begin{array}{c|cc} \text{sign} & D\ (u) & I\ (dv)\\\hline + & x^2 & \cos x\\ - & 2x & \sin x\\ + & 2 & -\cos x\\ - & 0 & -\sin x \end{array}

Multiply along diagonals with the signs:

∫x2cos⁡x dx=x2sin⁡x+2xcos⁡x−2sin⁡x+C.\int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x + C.

Cyclic parts. For ∫exsin⁡x dx\int e^x\sin x\,dx neither factor simplifies, but applying parts twice reproduces the original integral, which you solve for algebraically.

Example
Cyclic parts: ∫exsin⁡x dx\displaystyle\int e^{x}\sin x\,dx

Let I=∫exsin⁡x dxI=\int e^{x}\sin x\,dx. Take u=sin⁡x, dv=exdxu=\sin x,\ dv=e^x dx:  I=exsin⁡x−∫excos⁡x dx.\ I=e^x\sin x-\int e^x\cos x\,dx. Apply parts again to ∫excos⁡x dx\int e^x\cos x\,dx with u=cos⁡xu=\cos x:

I=exsin⁡x−(excos⁡x+∫exsin⁡x dx)I=exsin⁡x−excos⁡x−I2I=ex(sin⁡x−cos⁡x) ⇒ I=12ex(sin⁡x−cos⁡x)+C.\begin{aligned} I &= e^x\sin x-\Big(e^x\cos x+\int e^x\sin x\,dx\Big)\\ I &= e^x\sin x - e^x\cos x - I\\ 2I &= e^x(\sin x-\cos x)\ \Rightarrow\ I=\tfrac12 e^x(\sin x-\cos x)+C. \end{aligned}
Tip

Tip: For ∫ln⁡x dx\int \ln x\,dx and ∫arctan⁡x dx\int \arctan x\,dx there is no obvious dvdv — take dv=dxdv=dx (so v=xv=x) and let uu be the whole function. Result: ∫ln⁡x dx=xln⁡x−x+C\int\ln x\,dx=x\ln x-x+C.

Try it
Quick check

Find ∫xln⁡x dx\displaystyle\int x\ln x\,dx.

Show the answer

Answer: x22ln⁡x−x24+C\dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}+C. Logarithm comes first in LIATE, so u=ln⁡xu=\ln x, dv=x dxdv=x\,dx; then du=1x dxdu=\tfrac1x\,dx and v=x22v=\tfrac{x^2}{2}. The new integral is ∫x22⋅1x dx=x24\int\tfrac{x^2}{2}\cdot\tfrac1x\,dx=\tfrac{x^2}{4}.

Rational Functions by Partial Fractions

Concept
Decomposing a proper rational function

To integrate P(x)Q(x)\dfrac{P(x)}{Q(x)} with deg⁡P<deg⁡Q\deg P<\deg Q: factor Q(x)Q(x), then write the fraction as a sum of simpler pieces.

  • Each distinct linear factor (ax+b)(ax+b) contributes Aax+b\dfrac{A}{ax+b}.
  • A repeated linear factor (ax+b)k(ax+b)^k contributes A1ax+b+A2(ax+b)2+⋯+Ak(ax+b)k\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_k}{(ax+b)^k}.

If deg⁡P≥deg⁡Q\deg P\ge\deg Q, do polynomial long division first. Clear denominators, then solve for the constants (plug in convenient xx-values or match coefficients).

Example
Full partial-fraction integral (distinct linear factors)

Evaluate ∫5x−4x2−x−2 dx\displaystyle\int\frac{5x-4}{x^2-x-2}\,dx. Factor: x2−x−2=(x−2)(x+1)x^2-x-2=(x-2)(x+1).

5x−4(x−2)(x+1)=Ax−2+Bx+1  ⇒  5x−4=A(x+1)+B(x−2).\frac{5x-4}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1} \;\Rightarrow\; 5x-4=A(x+1)+B(x-2).

Let x=2x=2: 6=3A⇒A=26=3A\Rightarrow A=2. Let x=−1x=-1: −9=−3B⇒B=3-9=-3B\Rightarrow B=3. So

∫ ⁣(2x−2+3x+1)dx=2ln⁡∣x−2∣+3ln⁡∣x+1∣+C.\int\!\Big(\frac{2}{x-2}+\frac{3}{x+1}\Big)dx = 2\ln|x-2|+3\ln|x+1|+C.
Example
A repeated linear factor
3x+1(x+1)2=Ax+1+B(x+1)2,3x+1=A(x+1)+B.\frac{3x+1}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2},\quad 3x+1=A(x+1)+B.

x=−1: B=−2.x=-1:\ B=-2. Matching xx: A=3.A=3. Then

∫3x+1(x+1)2 dx=3ln⁡∣x+1∣+2x+1+C.\int\frac{3x+1}{(x+1)^2}\,dx=3\ln|x+1|+\frac{2}{x+1}+C.
Tip

Tip: Always check that the fraction is proper (deg⁡P<deg⁡Q\deg P<\deg Q) before decomposing. Otherwise divide first — the quotient integrates as a polynomial and the remainder gets the partial fractions.

Try it
Quick check

Find ∫dxx2−1\displaystyle\int\frac{dx}{x^2-1}.

Show the answer

Answer: 12ln⁡∣x−1x+1∣+C\dfrac12\ln\left|\dfrac{x-1}{x+1}\right|+C. 1(x−1)(x+1)=1/2x−1−1/2x+1\dfrac{1}{(x-1)(x+1)}=\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1} (cover-up at x=1x=1 and x=−1x=-1); integrate each term to a logarithm.

Trigonometric Integrals

Concept
Powers of sine and cosine

For ∫sin⁡mxcos⁡nx dx\displaystyle\int \sin^m x\cos^n x\,dx:

  • If a power is odd, peel off one factor and convert the rest with sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1, then substitute.
  • If both powers are even, use the power-reduction identities @@BLOCK0@@
Example
Odd power — substitute
∫sin⁡3x dx=∫(1−cos⁡2x)sin⁡x dx=u=cos⁡x−∫(1−u2) du=−cos⁡x+13cos⁡3x+C.\int\sin^3 x\,dx=\int(1-\cos^2 x)\sin x\,dx \overset{u=\cos x}{=}-\int(1-u^2)\,du=-\cos x+\tfrac13\cos^3 x+C.
Example
Even power — power reduction
∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−sin⁡2x4+C.\int\sin^2 x\,dx=\int\frac{1-\cos 2x}{2}\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.
Tip

Tip: The double-angle identity sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x lets you rewrite ∫sin⁡xcos⁡x dx=12∫sin⁡2x dx=−14cos⁡2x+C\int\sin x\cos x\,dx=\tfrac12\int\sin 2x\,dx=-\tfrac14\cos 2x+C instantly.

Try it
Quick check

Find ∫cos⁡3x dx\displaystyle\int\cos^3x\,dx.

Show the answer

Answer: sin⁡x−sin⁡3x3+C\sin x-\dfrac{\sin^3x}{3}+C. Odd power: save one cos⁡x\cos x and convert the rest, cos⁡3x=(1−sin⁡2x)cos⁡x\cos^3x=(1-\sin^2x)\cos x. With u=sin⁡xu=\sin x this is ∫(1−u2) du\int(1-u^2)\,du.

Trigonometric Substitution (a first look)

Concept
Which substitution to use

When a root of a quadratic appears, trade xx for a trig function so the Pythagorean identity collapses the root:

a2−x2: x=asin⁡θ,a2+x2: x=atan⁡θ,x2−a2: x=asec⁡θ.\sqrt{a^2-x^2}:\ x=a\sin\theta,\qquad \sqrt{a^2+x^2}:\ x=a\tan\theta,\qquad \sqrt{x^2-a^2}:\ x=a\sec\theta.

After substituting, integrate in θ\theta, then use a right triangle to convert back to xx.

Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Example
∫dx9−x2\displaystyle\int\frac{dx}{\sqrt{9-x^2}} with x=3sin⁡θx=3\sin\theta
θ3x√(9 − x²)
The substitution x=3sin⁡θx=3\sin\theta as a triangle: back-substitute by reading θ=arcsin⁡x3\theta=\arcsin\tfrac x3 off it.

Then dx=3cos⁡θ dθdx=3\cos\theta\,d\theta and 9−x2=3cos⁡θ\sqrt{9-x^2}=3\cos\theta, so

∫3cos⁡θ3cos⁡θ dθ=∫dθ=θ+C=arcsin⁡ ⁣(x3)+C.\int\frac{3\cos\theta}{3\cos\theta}\,d\theta=\int d\theta=\theta+C=\arcsin\!\Big(\frac{x}{3}\Big)+C.
Tip

Tip: Many a2−x2\sqrt{a^2-x^2} and a2+x2\sqrt{a^2+x^2} integrals are faster with the standard arcsin/arctan formulas. Reach for a full trig sub when the root is multiplied by other powers of xx.

Try it
Quick check

Evaluate ∫024−x2 dx\displaystyle\int_0^2\sqrt{4-x^2}\,dx with x=2sin⁡θx=2\sin\theta.

Show the answer

Answer: π\pi. dx=2cos⁡θ dθdx=2\cos\theta\,d\theta, 4−x2=2cos⁡θ\sqrt{4-x^2}=2\cos\theta, and the limits become 00 to π2\tfrac{\pi}{2}: ∫0π/24cos⁡2θ dθ=2∫0π/2(1+cos⁡2θ) dθ=π\displaystyle\int_0^{\pi/2}4\cos^2\theta\,d\theta=2\int_0^{\pi/2}(1+\cos2\theta)\,d\theta=\pi. It is a quarter of a circle of radius 22.

Improper Integrals

Concept
Rewrite as a limit

An integral is improper if a limit of integration is infinite, or if the integrand blows up somewhere on the interval. Handle it by replacing the bad endpoint with a limit:

∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx,∫abf(x) dx=lim⁡t→b−∫atf(x) dx  (vertical asymptote at b).\int_a^{\infty}f(x)\,dx=\lim_{t\to\infty}\int_a^{t}f(x)\,dx,\qquad \int_a^{b}f(x)\,dx=\lim_{t\to b^-}\int_a^{t}f(x)\,dx\ \ (\text{vertical asymptote at }b).

If the limit is a finite number, the integral converges; if it is ±∞\pm\infty or does not exist, it diverges. Split an integral with two trouble spots into two pieces — both must converge.

An integral over an infinite interval (or through a blow-up) is not a Riemann sum, so we define it as the limit of honest integrals over longer and longer intervals. Whether that limit is finite depends on how fast the integrand dies: 1x2\frac1{x^2} shrinks fast enough that the added pieces total a finite amount, while 1x\frac1x shrinks too slowly.

Example
Full improper integral (infinite limit) — converges
xy123456780.511/x (dashed)1/x²
The endless region under 1/x21/x^2 from 11 on has finite area 11; under 1/x1/x it would not.
∫1∞1x2 dx=lim⁡t→∞∫1tx−2 dx=lim⁡t→∞[−1x]1t=lim⁡t→∞(−1t+1)=0+1=1.(converges to 1)\begin{aligned} \int_1^{\infty}\frac{1}{x^2}\,dx &=\lim_{t\to\infty}\int_1^{t}x^{-2}\,dx =\lim_{t\to\infty}\Big[-\frac1x\Big]_1^{t}\\ &=\lim_{t\to\infty}\Big(-\frac1t+1\Big)=0+1=1.\quad(\textbf{converges to }1) \end{aligned}
Example
Unbounded integrand — diverges

1x→∞\dfrac1x\to\infty as x→0+x\to0^+, so

∫011x dx=lim⁡t→0+[ln⁡x]t1=lim⁡t→0+(0−ln⁡t)=+∞.(diverges)\int_0^{1}\frac{1}{x}\,dx=\lim_{t\to0^+}\big[\ln x\big]_t^{1} =\lim_{t\to0^+}(0-\ln t)=+\infty.\quad(\textbf{diverges})
Tip

pp-integral test: ∫1∞dxxp\displaystyle\int_1^{\infty}\frac{dx}{x^{p}} converges   ⟺  p>1\iff p>1;   ∫01dxxp\displaystyle\int_0^{1}\frac{dx}{x^{p}} converges   ⟺  p<1\iff p<1.

Try it
Quick check

Decide convergence and evaluate if possible: (a) ∫0∞e−2x dx\displaystyle\int_0^\infty e^{-2x}\,dx; (b) ∫1∞dxx\displaystyle\int_1^\infty\frac{dx}{\sqrt x}.

Show the answer

(a) Converges to 12\tfrac12: lim⁡b→∞[−12e−2x]0b=0+12\displaystyle\lim_{b\to\infty}\left[-\tfrac12e^{-2x}\right]_0^b=0+\tfrac12. (b) Diverges: lim⁡b→∞[2x]1b=∞\displaystyle\lim_{b\to\infty}\left[2\sqrt x\right]_1^b=\infty (a pp-integral with p=12≤1p=\tfrac12\le1).

Going Deeper: Advanced Techniques

Concept
1. Reduction formulas

Repeated integration by parts produces a formula that lowers a power by a fixed step. For example,

∫sin⁡nx dx=−sin⁡n−1xcos⁡xn+n−1n∫sin⁡n−2x dx.\int\sin^n x\,dx=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}\int\sin^{n-2}x\,dx.

Apply it repeatedly until you reach ∫sin⁡x dx\int\sin x\,dx or ∫dx\int dx. The same pattern gives reduction formulas for cos⁡nx\cos^n x, sec⁡nx\sec^n x, xnexx^n e^x, and (ln⁡x)n(\ln x)^n.

Example
2. Deriving cyclic parts in general

For ∫eaxsin⁡(bx) dx\int e^{ax}\sin(bx)\,dx, two applications of parts give

∫eaxsin⁡(bx) dx=eax(asin⁡bx−bcos⁡bx)a2+b2+C.\int e^{ax}\sin(bx)\,dx=\frac{e^{ax}\big(a\sin bx-b\cos bx\big)}{a^2+b^2}+C.

The denominator a2+b2a^2+b^2 is the fingerprint of a cyclic “solve-for-II” problem.

Example
3. Partial fractions with an irreducible quadratic

An irreducible factor (x2+px+q)(x^2+px+q) contributes a linear numerator Bx+Cx2+px+q\dfrac{Bx+C}{x^2+px+q}. Evaluate:

2x+1x(x2+1)=Ax+Bx+Cx2+1.\frac{2x+1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}.

x=0: A=1.x=0:\ A=1. Matching: A+B=0⇒B=−1A+B=0\Rightarrow B=-1, and C=2C=2. So

∫2x+1x(x2+1) dx=ln⁡∣x∣−12ln⁡(x2+1)+2arctan⁡x+C,\int\frac{2x+1}{x(x^2+1)}\,dx=\ln|x|-\tfrac12\ln(x^2+1)+2\arctan x+C,

using ∫xx2+1dx=12ln⁡(x2+1)\int\frac{x}{x^2+1}dx=\tfrac12\ln(x^2+1) and ∫1x2+1dx=arctan⁡x\int\frac{1}{x^2+1}dx=\arctan x.

Concept
4. The comparison test for improper integrals

If 0≤f(x)≤g(x)0\le f(x)\le g(x) for all x≥ax\ge a, then:

∫a∞g dx converges ⇒ ∫a∞f dx converges;∫a∞f dx diverges ⇒ ∫a∞g dx diverges.\int_a^{\infty}g\,dx\ \text{converges}\ \Rightarrow\ \int_a^{\infty}f\,dx\ \text{converges};\qquad \int_a^{\infty}f\,dx\ \text{diverges}\ \Rightarrow\ \int_a^{\infty}g\,dx\ \text{diverges}.

Use it when an antiderivative is hard to find but the integrand is squeezed by a pp-integral.

Example
5. Comparison test in action

Does ∫1∞1x2+x dx\displaystyle\int_1^{\infty}\frac{1}{x^2+x}\,dx converge? For x≥1x\ge1,  x2+x>x2>0\ x^2+x> x^2>0, so 0<1x2+x<1x20<\frac{1}{x^2+x}<\frac{1}{x^2}. Since ∫1∞dxx2\int_1^{\infty}\frac{dx}{x^2} converges (a pp-integral with p=2>1p=2>1), the smaller integral converges by comparison.

Example
6. A full trig substitution with a leftover power

Evaluate ∫x24−x2 dx\displaystyle\int\frac{x^2}{\sqrt{4-x^2}}\,dx with x=2sin⁡θx=2\sin\theta, dx=2cos⁡θ dθdx=2\cos\theta\,d\theta, 4−x2=2cos⁡θ\sqrt{4-x^2}=2\cos\theta:

∫4sin⁡2θ2cos⁡θ 2cos⁡θ dθ=∫4sin⁡2θ dθ=2∫(1−cos⁡2θ) dθ=2θ−sin⁡2θ+C=2θ−2sin⁡θcos⁡θ+C=2arcsin⁡ ⁣(x2)−x4−x22+C,\begin{aligned} \int\frac{4\sin^2\theta}{2\cos\theta}\,2\cos\theta\,d\theta &=\int 4\sin^2\theta\,d\theta=2\int(1-\cos2\theta)\,d\theta\\ &=2\theta-\sin2\theta+C=2\theta-2\sin\theta\cos\theta+C\\ &=2\arcsin\!\Big(\frac{x}{2}\Big)-\frac{x\sqrt{4-x^2}}{2}+C, \end{aligned}

reading sin⁡θ=x2, cos⁡θ=4−x22\sin\theta=\frac{x}{2},\ \cos\theta=\frac{\sqrt{4-x^2}}{2} off the reference triangle.

Concept
7. Completing the square before substituting

When a quadratic has no rational roots, complete the square to expose a standard form. For ∫dxx2+2x+5\int\frac{dx}{x^2+2x+5}, write x2+2x+5=(x+1)2+4x^2+2x+5=(x+1)^2+4, then let u=x+1u=x+1:

∫duu2+4=12arctan⁡ ⁣(u2)+C=12arctan⁡ ⁣(x+12)+C.\int\frac{du}{u^2+4}=\frac12\arctan\!\Big(\frac{u}{2}\Big)+C=\frac12\arctan\!\Big(\frac{x+1}{2}\Big)+C.
Tip

Big picture: Before integrating, run a quick checklist — (1) does a simple uu-sub work? (2) is it a product suited to parts? (3) is it a rational function (partial fractions)? (4) powers of trig? (5) a root of a quadratic (trig sub / complete the square)? Matching the form to the method is the whole game.

Try it
Quick check

∫ln⁡x dx\displaystyle\int \ln x\,dx

Work it on paper before revealing the answer.

Show the answer

Answer: xln⁡x−x+Cx\ln x-x+C

Why It Works

Integration by parts is the product rule integrated: (uv)′=u′v+uv′(uv)' = u'v + uv', so ∫uv′=uv−∫u′v\int uv' = uv - \int u'v. Substitution is the chain rule integrated: since ddxF(g(x))=f(g(x))g′(x)\tfrac{d}{dx}F(g(x)) = f(g(x))g'(x), we get ∫f(g(x))g′(x) dx=F(g(x))+C\int f(g(x))g'(x)\,dx = F(g(x)) + C. Every integration technique is a differentiation rule read from right to left — which is why differentiating your answer is always a complete check.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Substitution and integration by parts
∫f(g(x))g′(x) dx=∫f(u) du,∫u dv=uv−∫v du\int f(g(x))g'(x)\,dx=\int f(u)\,du, \qquad \int u\,dv = uv-\int v\,du

What it means. Substitution reverses the chain rule; parts reverses the product rule.

Example. ∫2x ex2 dx=ex2+C\int 2x\,e^{x^2}\,dx=e^{x^2}+C (let u=x2u=x^2, du=2x dxdu=2x\,dx).

Why it works. With u=g(x)u=g(x), du=g′(x)dxdu=g'(x)dx, so the integral literally rewrites in uu. For parts, integrate the product rule (uv)′=u′v+uv′(uv)'=u'v+uv' across the interval and rearrange.

Tip. When substituting in a definite integral, convert the limits to uu as well — then you never convert back.