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Worked Examples · Example 4

Q.Evaluate the following integrals by the method of substitution:

(a) ∫xx+2 dx\int x\sqrt{x+2}\,dx
(b) ∫xx−1 dx\int \frac{x}{\sqrt{x-1}}\,dx
(c) ∫33x 3x dx\int 3^{3^x}\,3^x\,dx
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
7% · 4/59 Questions
✓ Free question

Substitute to remove the radical/exponent, integrate as powers or an exponential, then back-substitute.

∫xndx=xn+1n+1+C,∫au du=aulog⁡a+C.\int x^n dx=\frac{x^{n+1}}{n+1}+C,\qquad \int a^{u}\,du=\frac{a^{u}}{\log a}+C.

Steps

  1. (a) Let u=x+2⇒x=u−2, dx=duu=x+2\Rightarrow x=u-2,\ dx=du:

∫(u−2)u1/2du=∫ ⁣(u3/2−2u1/2)du=25u5/2−43u3/2+C.\int(u-2)u^{1/2}du=\int\!\big(u^{3/2}-2u^{1/2}\big)du=\frac{2}{5}u^{5/2}-\frac{4}{3}u^{3/2}+C.

Back-substitute: 25(x+2)5/2−43(x+2)3/2+C.\dfrac{2}{5}(x+2)^{5/2}-\dfrac{4}{3}(x+2)^{3/2}+C.

  1. (b) Let u=x−1⇒x=u+1, dx=duu=x-1\Rightarrow x=u+1,\ dx=du:

∫u+1u du=∫ ⁣(u1/2+u−1/2)du=23u3/2+2u1/2+C.\int\frac{u+1}{\sqrt{u}}\,du=\int\!\big(u^{1/2}+u^{-1/2}\big)du=\frac{2}{3}u^{3/2}+2u^{1/2}+C.

Back-substitute: 23(x−1)3/2+2x−1+C.\dfrac{2}{3}(x-1)^{3/2}+2\sqrt{x-1}+C.

  1. (c) Let u=3x⇒du=3xlog⁡3 dxu=3^{x}\Rightarrow du=3^{x}\log 3\,dx, so 3xdx=dulog⁡33^{x}dx=\dfrac{du}{\log 3}:

∫33x 3x dx=1log⁡3∫3u du=1log⁡3⋅3ulog⁡3=33x(log⁡3)2+C.\int 3^{3^{x}}\,3^{x}\,dx=\frac{1}{\log 3}\int 3^{u}\,du=\frac{1}{\log 3}\cdot\frac{3^{u}}{\log 3}=\frac{3^{3^{x}}}{(\log 3)^2}+C.

✓Final answer

  1. 25(x+2)5/2−43(x+2)3/2+C\dfrac{2}{5}(x+2)^{5/2}-\dfrac{4}{3}(x+2)^{3/2}+C;
  2. 23(x−1)3/2+2x−1+C\dfrac{2}{3}(x-1)^{3/2}+2\sqrt{x-1}+C;
  3. 33x(log⁡3)2+C\dfrac{3^{3^{x}}}{(\log 3)^2}+C.

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