Q.Evaluate ∫(3x+2)4dx using substitution.
Concept understanding — Integration by Substitution
The substitution (change-of-variable) method mirrors the chain rule of differentiation. If u=g(x) is a differentiable function, then
∫f(g(x))g′(x)dx=∫f(u)du,
because du=g′(x)dx. Choosing u so that its derivative already appears (up to a constant) in the integrand converts a hard integral into a standard one; after integrating in u, substitute back u=g(x).
Two especially useful consequences (with u=f(x)):
∫f(x)f′(x)dx=log∣f(x)∣+c,∫f′(x)[f(x)]ndx=n+1[f(x)]n+1+c (n=−1).
Standard log-form results that follow are ∫tanxdx=log∣secx∣+c, ∫cotxdx=log∣sinx∣+c, ∫cosecxdx=log∣cosecx−cotx∣+c, and ∫secxdx=log∣secx+tanx∣+c.
For a trigonometric substitution (e.g. x=atanθ), draw a right triangle to read back the other trig ratios when reversing the substitution.
The whole method rests on picking a u whose differential g′(x)dx is present in the integrand. If it isn't (even up to a constant multiple), substitution won't simplify things — try a different method.
The same substitution technique applies to any linear-inner-expression power, regardless of the exponent's size.
15(3x+2)5+C.
Let u=3x+2, du=3dx; ∫(3x+2)4dx=31∫u4du=15u5+C.
15(3x+2)5+C
Substituting
Let u=3x+2, so du=3dx, i.e. dx=3du.
∫(3x+2)4dx=∫u4⋅3du=31⋅5u5+C=15u5+C=15(3x+2)5+C
Check (differentiate the answer back, using the chain rule): dxd[15(3x+2)5]=155(3x+2)4⋅3=(3x+2)4 — exactly the original integrand.
15(3x+2)5+C
Forgetting to include the 31 factor arising from du=3dx, which would leave the final answer three times too large.
- CA Foundation 2025Set jan-20251 markMCQQ.∫(2x+5)7dx (A) 16(2x+5)8 (B) 7(2x+5)7 (C) 2(2x2+5x)7 (D) 5(2x2+5x)7
›Reveal solutionSolution
∫(2x+5)7dx=2⋅8(2x+5)8=16(2x+5)8+C.
Step 1 — Use the standard form for a linear substitution
∫(ax+b)ndx=a(n+1)(ax+b)n+1+C
Step 2 — Identify the constants
a=2,b=5,n=7
Step 3 — Substitute
∫(2x+5)7dx=2×8(2x+5)8=16(2x+5)8+C
Why the other options are wrong: (B) divides only by 7 (raises to the wrong power AND drops the a1); (C) and (D) wrongly expand (2x+5) as (2x2+5x), which is not the inner function.
Watch outDivide by BOTH the new power (n+1=8) and the coefficient of x (a=2) → denominator 16. Missing the a1 factor is the most common mistake.
TipVerify by differentiating: dxd16(2x+5)8=168(2x+5)7⋅2=(2x+5)7. ✓
✓Final answer(A) 16(2x+5)8
- CA Foundation 2024Set sep-20241 markMCQQ.Evaluate the following integral ∫x(x5+1)1dx. (A) log(x5+1x5)+c (B) 51log(x5+1x5)+c (C) 31log(x5+1x5)+c (D) 31log(x5x5+1)+c
›Reveal solutionSolution
Substitute u = x⁵ then use partial fractions: the integral = (1/5)·ln(x⁵/(x⁵+1)) + c.
Step 1 — Prepare for substitution
Multiply top and bottom by x4:
∫x(x5+1)1dx=∫x5(x5+1)x4dx
Step 2 — Substitute u = x⁵
With u=x5, du=5x4dx, i.e. x4dx=5du:
=51∫u(u+1)du
Step 3 — Partial fractions and integrate
u(u+1)1=u1−u+11
51∫(u1−u+11)du=51[lnu−ln(u+1)]+c=51ln(x5+1x5)+c
Why the other options are wrong: (A) omits the 1/5 factor; (C) uses 1/3 (wrong power); (D) inverts the ratio to (x⁵+1)/x⁵. Only (B) has both the 1/5 and the correct x⁵/(x⁵+1).
Watch outThe constant is 1/5 (because du = 5x⁴dx introduces the 5), and the ratio inside the log is x⁵/(x⁵+1) — coming from ln u − ln(u+1), not the reverse.
TipFor ∫1/[x(xⁿ+1)]dx, multiply by xⁿ⁻¹/xⁿ⁻¹ and set u = xⁿ; the answer is always (1/n)·ln(xⁿ/(xⁿ+1)) + c.
✓Final answer(B) 51log(x5+1x5)+c
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