MHT-CET Maths · Indefinite Integration

Integration by Substitution — the Workhorse Method

Spot an inner function whose derivative also appears in the integrand, substitute to rename it, and the integral collapses to a standard form.

Why this matters

44 PYQs — by far the largest bucket in the chapter, and the method every other technique falls back on. The single most-tested pattern is f'(x)/f(x) → log|f(x)|. Beyond that: powers of a function times its derivative, root substitutions, and exponential substitutions. Difficulty is steep here (about 60% HARD), but every one of them reduces to 'find u, find du, rewrite, integrate'.

Concept 1 of 5

The Substitution Rule

Intuition

If the integrand contains some inner function g(x)g(x) AND a factor that looks like its derivative g(x)g'(x), substitute u=g(x)u = g(x). The g(x)dxg'(x)\,dx becomes dudu, and the whole integral turns into something standard in uu.

Definition

If u=g(x)u = g(x), then du=g(x)dxdu = g'(x)\,dx, and f(g(x))g(x)dx=f(u)du\int f(g(x))\,g'(x)\,dx = \int f(u)\,du. The art is choosing g(x)g(x) so that its derivative is already present (up to a constant) in the integrand.

Substitution rule

f(g(x))g(x)dx=f(u)du,u=g(x)\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\quad u = g(x)
  • u = g(x)the inner function you rename
  • du = g'(x)\,dxits differential, which must appear in the integrand

Worked example

Evaluate sin(1/x)x2dx\displaystyle\int \dfrac{\sin(1/x)}{x^2}\,dx.
  1. Let u=1xu = \dfrac{1}{x}. Then du=1x2dxdu = -\dfrac{1}{x^2}\,dx, so dxx2=du\dfrac{dx}{x^2} = -\,du.
  2. Rewrite: sinu(du)=sinudu\int \sin u \,(-du) = -\int \sin u\,du.
  3. Standard form: (cosu)+C=cosu+C-(-\cos u) + C = \cos u + C.
  4. Back-substitute u=1/xu = 1/x.
Answer:cos1x+C\cos\dfrac{1}{x} + C
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate (logx)2xdx\displaystyle\int \dfrac{(\log x)^2}{x}\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    2xex2dx\int 2x\,e^{x^2}\,dx
  2. 2.
    cosxesinxdx\int \cos x\,e^{\sin x}\,dx
  3. 3.
    (logx)xdx\int \dfrac{(\log x)}{x}\,dx
  4. 4.
    sinxcosxdx\int \sin x\,\cos x\,dx

From the bank · past-year question

Example 1Indefinite IntegrationMODERATE
tan(1/x)x2dx=\int\frac{\tan(1/x)}{x^2}\,dx =

[Q125 · 13th May Shift 1 · 2024]

Adjust for the missing constant

If du=g(x)dxdu = g'(x)\,dx appears up to a numeric factor (e.g. you have xdxx\,dx but du=2xdxdu = 2x\,dx), pull the constant out: xdx=12dux\,dx = \tfrac12\,du. Forgetting the 12\tfrac12 is the most common slip.

Concept 2 of 5

The f'(x)/f(x) → log Pattern

Intuition

Whenever the numerator is exactly the derivative of the denominator, the integral is just log\log of the denominator. This is the single highest-yield pattern in the chapter — train your eye to spot 'top = derivative of bottom'.

Definition

If the integrand is a fraction whose numerator is the derivative of its denominator, then f(x)f(x)dx=logf(x)+C\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C. Often you must engineer the numerator: split it into 'a constant times f(x)f'(x)' plus a leftover, then integrate each piece.

Logarithmic integral

f(x)f(x)dx=logf(x)+C\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C
  • f(x)the denominator
  • f'(x)its derivative — must equal the numerator (up to a constant)

Worked example

Evaluate 2x+3x2+3x+5dx\displaystyle\int \dfrac{2x + 3}{x^2 + 3x + 5}\,dx.
  1. The denominator is f=x2+3x+5f = x^2 + 3x + 5; its derivative is f=2x+3f' = 2x + 3 — exactly the numerator.
  2. So the integrand has the form f/ff'/f, which integrates to logf\log|f|.
  3. 2x+3x2+3x+5dx=logx2+3x+5+C\int \dfrac{2x+3}{x^2+3x+5}\,dx = \log|x^2+3x+5| + C.
Answer:logx2+3x+5+C\log|x^2 + 3x + 5| + C
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate 3x2+2xx3+x2+1dx\displaystyle\int \dfrac{3x^2 + 2x}{x^3 + x^2 + 1}\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    2xx2+1dx\int \dfrac{2x}{x^2+1}\,dx
  2. 2.
    tanxdx\int \tan x\,dx
  3. 3.
    exex+1dx\int \dfrac{e^x}{e^x+1}\,dx
  4. 4.
    cotxdx\int \cot x\,dx

From the bank · past-year question

Example 2Indefinite IntegrationHARD
If 4ex252ex5dx=Ax+Blog2ex5+c\int\frac{4e^x-25}{2e^x-5}dx = Ax+B\log|2e^x-5|+c (where c is a constant of integration) then

[Q123 · 4th May Shift 2 · 2023]

Engineer the numerator into f'(x) + leftover

Rarely is the top exactly f(x)f'(x). Write it as 'constant ×f(x)\times f'(x) + remainder', send the first piece to a clean log, and handle the remainder separately. The whole-number coefficient comes from matching.

Concept 3 of 5

Power of a Function times its Derivative

Intuition

When you see some function raised to a power, multiplied by (a constant times) its derivative, substitution turns it into the plain power rule. The denominator-squared shapes f/f2\int f'/f^2 are the same idea with a negative power.

Definition

If u=f(x)u = f(x) then [f(x)]nf(x)dx=[f(x)]n+1n+1+C\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C for n1n \neq -1 (and the n=1n = -1 case is the log pattern). The shape f(x)[f(x)]2dx=1f(x)+C\displaystyle\int \dfrac{f'(x)}{[f(x)]^2}\,dx = -\dfrac{1}{f(x)} + C is the n=2n = -2 instance.

Power-of-a-function rule

[f(x)]nf(x)dx=[f(x)]n+1n+1+C(n1)\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)

Worked example

Evaluate (x3+2x+1)4(3x2+2)dx\displaystyle\int (x^3 + 2x + 1)^4\,(3x^2 + 2)\,dx.
  1. Let u=x3+2x+1u = x^3 + 2x + 1. Then du=(3x2+2)dxdu = (3x^2 + 2)\,dx — exactly the second factor.
  2. Rewrite: u4du=u55+C\int u^4\,du = \dfrac{u^5}{5} + C.
  3. Back-substitute.
Answer:(x3+2x+1)55+C\dfrac{(x^3 + 2x + 1)^5}{5} + C
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate (x2+1)3(2x)dx\displaystyle\int (x^2+1)^3\,(2x)\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    (1+x2)42xdx\int (1+x^2)^4\,2x\,dx
  2. 2.
    sec2xtan2xdx\int \dfrac{\sec^2 x}{\tan^2 x}\,dx
  3. 3.
    sin3xcosxdx\int \sin^3 x\,\cos x\,dx
  4. 4.
    cosxsin2xdx\int \dfrac{\cos x}{\sin^2 x}\,dx

From the bank · past-year question

Example 3Indefinite IntegrationMODERATE
2+cosx(2x+sinx)2dx=\int\frac{2+\cos x}{(2x+\sin x)^2}\,dx=

[Q103 · 13th May Shift 2 · 2024]

n = −1 is NOT this rule

If the power is 1-1 (i.e. f/f\int f'/f), the power rule blows up. That case is the log pattern. Every other integer/fraction power uses un+1n+1\dfrac{u^{n+1}}{n+1}.

Concept 4 of 5

Root and Linear-Radical Substitutions

Intuition

A square root of a linear (or simple) expression is cleared by substituting the root itself, or the inside of the root, as uu. The substitution removes the radical and leaves a polynomial or standard form.

Definition

For integrals containing ax+b\sqrt{ax+b}, substitute t=ax+bt = \sqrt{ax+b} (so xx and dxdx are expressed via tt), or substitute u=ax+bu = ax+b. For x\sqrt{x} shapes, t=xt = \sqrt{x} gives x=t2, dx=2tdtx = t^2,\ dx = 2t\,dt. The radical disappears and the integrand becomes rational/polynomial in tt.

Linear-radical substitution

t=ax+b  x=t2ba,dx=2tadtt = \sqrt{ax+b}\ \Rightarrow\ x = \dfrac{t^2 - b}{a},\quad dx = \dfrac{2t}{a}\,dt

Worked example

Evaluate 11+xdx\displaystyle\int \dfrac{1}{1 + \sqrt{x}}\,dx.
  1. Let t=xt = \sqrt{x}, so x=t2x = t^2 and dx=2tdtdx = 2t\,dt.
  2. Rewrite: 11+t(2tdt)=2t1+tdt=2 ⁣(111+t)dt\int \dfrac{1}{1+t}\,(2t\,dt) = 2\int \dfrac{t}{1+t}\,dt = 2\int\!\left(1 - \dfrac{1}{1+t}\right)dt.
  3. Integrate: 2(tlog1+t)+C2\big(t - \log|1+t|\big) + C.
  4. Back-substitute t=xt = \sqrt{x}.
Answer:2x2log ⁣1+x+C2\sqrt{x} - 2\log\!\left|1 + \sqrt{x}\right| + C
Practice this conceptself-check · 4 quick reps

Try it yourself

If (2x+4)x1dx=a(x1)5/2+b(x1)3/2+c\displaystyle\int (2x+4)\sqrt{x-1}\,dx = a(x-1)^{5/2} + b(x-1)^{3/2} + c, find a+ba + b.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    dxx+1\int \dfrac{dx}{\sqrt{x}+1} — substitution?
  2. 2.
    2x+1dx\int \sqrt{2x+1}\,dx
  3. 3.
    xx1dx\int x\sqrt{x-1}\,dx — substitution?
  4. 4.
    dxx+3\int \dfrac{dx}{\sqrt{x+3}}

From the bank · past-year question

Example 4Indefinite IntegrationMODERATE
xx+1dx=\int\frac{\sqrt{x}}{x+1}\,dx=

[Q131 · 10th May Shift 1 · 2023]

Re-express EVERY x, including dx

After t=xt = \sqrt{x}, both the integrand AND dx=2tdtdx = 2t\,dt must be rewritten. Leaving a stray xx or the old dxdx behind is the classic substitution error.

Concept 5 of 5

Exponential and Special Substitutions

Intuition

When an exponential is buried inside a root or a tower of exponents, substitute the exponential itself. t=ext = e^x (so dt=exdxdt = e^x dx) clears most exe^x integrals; nested towers like 33x3^{3^x} substitute the inner power.

Definition

For integrands built from exe^x: substitute t=ext = e^x, giving dt=exdxdt = e^x\,dx; a stray exe^x in the numerator becomes dtdt and the rest becomes rational in tt. For exponential towers aaxaxdx\int a^{a^x}\,a^x\,dx, substitute u=axu = a^x (then du=axlogadxdu = a^x \log a\,dx).

Exponential substitution

t=ex  dt=exdxt = e^{x}\ \Rightarrow\ dt = e^{x}\,dx

Worked example

Evaluate ex1+e2xdx\displaystyle\int \dfrac{e^x}{1 + e^{2x}}\,dx.
  1. Let t=ext = e^x, so dt=exdxdt = e^x\,dx. Note e2x=(ex)2=t2e^{2x} = (e^x)^2 = t^2.
  2. Rewrite: dt1+t2=tan1t+C\int \dfrac{dt}{1 + t^2} = \tan^{-1}t + C.
  3. Back-substitute t=ext = e^x.
Answer:tan1(ex)+C\tan^{-1}(e^x) + C
Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate 33x3xdx\displaystyle\int 3^{3^x}\cdot 3^x\,dx.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    exex+1dx\int \dfrac{e^x}{e^x+1}\,dx
  2. 2.
    ex(1+ex)3dx\int e^x(1+e^x)^3\,dx
  3. 3.
    22x2xdx\int 2^{2^x}\,2^x\,dx
  4. 4.
    dx1+ex\int \dfrac{dx}{1+e^{-x}}

From the bank · past-year question

Example 5Indefinite IntegrationMODERATE
ex1dx=\int\sqrt{e^x - 1}\,dx =

[Q119 · 9th May Shift 2 · 2023]

Towers: substitute the INNER exponential

For aaxaxdx\int a^{a^x} a^x\,dx, the right substitution is u=axu = a^x, not u=aaxu = a^{a^x}. The stray axdxa^x\,dx is what becomes dudu (up to loga\log a).

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

  • The Substitution Rule

    Substitution rule

    f(g(x))g(x)dx=f(u)du,u=g(x)\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\quad u = g(x)
  • The f'(x)/f(x) → log Pattern

    Logarithmic integral

    f(x)f(x)dx=logf(x)+C\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C
  • Power of a Function times its Derivative

    Power-of-a-function rule

    [f(x)]nf(x)dx=[f(x)]n+1n+1+C(n1)\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)
  • Root and Linear-Radical Substitutions

    Linear-radical substitution

    t=ax+b  x=t2ba,dx=2tadtt = \sqrt{ax+b}\ \Rightarrow\ x = \dfrac{t^2 - b}{a},\quad dx = \dfrac{2t}{a}\,dt
  • Exponential and Special Substitutions

    Exponential substitution

    t=ex  dt=exdxt = e^{x}\ \Rightarrow\ dt = e^{x}\,dx

Watch out for (5)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Indefinite IntegrationHARD
If log(t+1+t2)1+t2dt=12[g(t)]2+c\int\frac{\log(t+\sqrt{1+t^2})}{\sqrt{1+t^2}}\,dt=\frac{1}{2}[g(t)]^2+c, (Where cc is a constant of integration) then g(2)g(2) is:

[Q129 · 13th May Shift 2 · 2024]

Example 2Indefinite IntegrationHARD
If cos3/5xsin3xdx=1mcosmx+1ncosnx+c\displaystyle\int\cos^{3/5}x\cdot\sin^{3}x\,dx=-\frac{1}{m}\cos^{m}x+\frac{1}{n}\cos^{n}x+c (where cc is the constant of integration), then (m,n)=(m,n)=

[Q149 · 12th May Shift 1 · 2024]

Example 3Indefinite IntegrationMODERATE
If x+12x1dx=f(x)2x1+c\int \frac{x+1}{\sqrt{2x-1}}\,dx = f(x)\sqrt{2x-1} + c, then f(x)f(x) is equal to

[Q106 · 2nd May Shift 2 · 2023]

Example 4Indefinite IntegrationMODERATE
If dxex1=2tan1(f(x))+c\int \frac{dx}{e^x - 1} = 2\tan^{-1}(f(x)) + c where x>0x > 0, then f(x)f(x) is

[Q128 · 10th May Shift 2 · 2023]

Example 5Indefinite IntegrationHARD
The value of (x21)dxx32x42x2+1\displaystyle\int\frac{(x^2-1)\,dx}{x^3\sqrt{2x^4-2x^2+1}} is

[Q113 · 10th May Shift 2 · 2024]

Drill every past-year question on this subtopic

44 questions from the bank — paginated, with cart and Word-export support.

Related notes