Q.Evaluate ∫(2x+3)5dx 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.
Substituting u for the inner linear expression converts this into a standard power-rule integral.
12(2x+3)6+C.
Let u=2x+3, du=2dx; ∫(2x+3)5dx=21∫u5du=12u6+C.
12(2x+3)6+C
Substituting
Let u=2x+3, so du=2dx, i.e. dx=2du.
∫(2x+3)5dx=∫u5⋅2du=21∫u5du=21⋅6u6+C=12u6+C=12(2x+3)6+C
Check (differentiate the answer back, using the chain rule): dxd[12(2x+3)6]=126(2x+3)5⋅2=(2x+3)5 — exactly the original integrand.
12(2x+3)6+C
Forgetting to divide by the derivative of the inner expression (here, forgetting the 21 from du=2dx), which would leave the answer off by a constant factor.
- 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
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.