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3.1 · Q2

Q.Evaluate the following by substitution method:

(i) ∫x+e2xx2+e2x dx\int \frac{x+e^{2x}}{x^2+e^{2x}}\,dx
(ii) ∫dxx+x\int \frac{dx}{\sqrt{x}+x}
(iii) ∫ex(1+x)(1+xex)2 dx\int \frac{e^x(1+x)}{(1+xe^x)^2}\,dx
(iv) ∫2xx2+13 dx\int \frac{2x}{\sqrt[3]{x^2+1}}\,dx
(v) ∫e2x+e−2xe2x−e−2x dx\int \frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\,dx
(vi) ∫3ex−5e−x4ex+5e−x dx\int \frac{3e^x-5e^{-x}}{4e^x+5e^{-x}}\,dx
(vii) ∫2x−3x2−3x−18 dx\int \frac{2x-3}{x^2-3x-18}\,dx
(viii) ∫1x(1+log⁡x)2 dx\int \frac{1}{x(1+\log x)^2}\,dx
(ix) ∫ax−1log⁡a+xa−1ax+xa dx\int \frac{a^{x-1}\log a + x^{a-1}}{a^x+x^a}\,dx
Lakshadweep CbseNCERTSubjective· 5mImportance★★★★★
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Nine substitution integrals; most are of the form ∫g′(x)g(x)dx=log⁡∣g(x)∣\int \frac{g'(x)}{g(x)}dx=\log|g(x)| or ∫undu\int u^n du.

∫g′(x)g(x) dx=log⁡∣g(x)∣+C\displaystyle\int \frac{g'(x)}{g(x)}\,dx=\log|g(x)|+C; ∫un du=un+1n+1+C\displaystyle\int u^n\,du=\frac{u^{n+1}}{n+1}+C; and ∫duu2=−1u+C\displaystyle\int\frac{du}{u^2}=-\frac1u+C.

(i) ∫x+e2xx2+e2x dx\displaystyle\int \frac{x+e^{2x}}{x^2+e^{2x}}\,dx

  1. ddx(x2+e2x)=2x+2e2x=2(x+e2x)\dfrac{d}{dx}(x^2+e^{2x})=2x+2e^{2x}=2(x+e^{2x}), so numerator =12×=\tfrac12\times(denominator)′'.
  2. =12log⁡∣x2+e2x∣+C.=\tfrac12\log|x^2+e^{2x}|+C.

(ii) ∫dxx+x=∫dxx(1+x)\displaystyle\int \frac{dx}{\sqrt{x}+x}=\int\frac{dx}{\sqrt{x}(1+\sqrt{x})}

3. Put u=x⇒du=12xdx⇒dx=2u du.u=\sqrt{x}\Rightarrow du=\tfrac{1}{2\sqrt{x}}dx\Rightarrow dx=2u\,du.

4. =∫2u duu(1+u)=2∫du1+u=2log⁡(1+x)+C.=\displaystyle\int\frac{2u\,du}{u(1+u)}=2\int\frac{du}{1+u}=2\log(1+\sqrt{x})+C.

(iii) ∫ex(1+x)(1+xex)2 dx\displaystyle\int \frac{e^x(1+x)}{(1+xe^x)^2}\,dx

5. Put u=1+xex⇒du=(ex+xex)dx=ex(1+x)dx.u=1+xe^x\Rightarrow du=(e^x+xe^x)dx=e^x(1+x)dx.

6. =∫duu2=−1u=−11+xex+C.=\displaystyle\int\frac{du}{u^2}=-\frac{1}{u}=-\frac{1}{1+xe^x}+C.

(iv) ∫2xx2+13 dx\displaystyle\int \frac{2x}{\sqrt[3]{x^2+1}}\,dx

7. Put u=x2+1⇒du=2x dx.u=x^2+1\Rightarrow du=2x\,dx.

8. =∫u−1/3du=u2/32/3=32(x2+1)2/3+C.=\displaystyle\int u^{-1/3}du=\frac{u^{2/3}}{2/3}=\frac32(x^2+1)^{2/3}+C.

(v) ∫e2x+e−2xe2x−e−2x dx\displaystyle\int \frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\,dx

9. Put u=e2x−e−2x⇒du=2(e2x+e−2x)dx.u=e^{2x}-e^{-2x}\Rightarrow du=2(e^{2x}+e^{-2x})dx.

10. =12∫duu=12log⁡∣e2x−e−2x∣+C.=\tfrac12\displaystyle\int\frac{du}{u}=\tfrac12\log|e^{2x}-e^{-2x}|+C.

(vi) ∫3ex−5e−x4ex+5e−x dx\displaystyle\int \frac{3e^x-5e^{-x}}{4e^x+5e^{-x}}\,dx

11. Write numerator =A(4ex+5e−x)+B(4ex−5e−x)=A(4e^x+5e^{-x})+B(4e^x-5e^{-x}) (denominator and its derivative).

12. Match: exe^x: 4A+4B=34A+4B=3; e−xe^{-x}: 5A−5B=−5⇒A−B=−1.5A-5B=-5\Rightarrow A-B=-1. Solving: A=−18, B=78.A=-\tfrac18,\ B=\tfrac78.

13. =Ax+Blog⁡∣4ex+5e−x∣=−18x+78log⁡∣4ex+5e−x∣+C.=Ax+B\log|4e^x+5e^{-x}|=-\tfrac18 x+\tfrac78\log|4e^x+5e^{-x}|+C.

(vii) ∫2x−3x2−3x−18 dx\displaystyle\int \frac{2x-3}{x^2-3x-18}\,dx

14. (x2−3x−18)′=2x−3=(x^2-3x-18)'=2x-3= numerator, so =log⁡∣x2−3x−18∣+C.=\log|x^2-3x-18|+C.

(viii) ∫1x(1+log⁡x)2 dx\displaystyle\int \frac{1}{x(1+\log x)^2}\,dx

15. Put u=1+log⁡x⇒du=1xdx.u=1+\log x\Rightarrow du=\tfrac{1}{x}dx.

16. =∫duu2=−1u=−11+log⁡x+C.=\displaystyle\int\frac{du}{u^2}=-\frac1u=-\frac{1}{1+\log x}+C.

(ix) ∫ax−1log⁡a+xa−1ax+xa dx\displaystyle\int \frac{a^{x-1}\log a + x^{a-1}}{a^x+x^a}\,dx

17. ddx(ax+xa)=axlog⁡a+a xa−1.\dfrac{d}{dx}(a^x+x^a)=a^x\log a+ a\,x^{a-1}. Numerator =ax−1log⁡a+xa−1=1a(axlog⁡a+a xa−1)=1a×=a^{x-1}\log a+x^{a-1}=\tfrac1a\big(a^x\log a+a\,x^{a-1}\big)=\tfrac1a\times(denominator)′'.

18. =1alog⁡∣ax+xa∣+C.=\dfrac{1}{a}\log|a^x+x^a|+C.

✓Final answer

(i) 12log⁡∣x2+e2x∣+C\tfrac12\log|x^2+e^{2x}|+C;

(ii) 2log⁡(1+x)+C2\log(1+\sqrt{x})+C;

(iii) −11+xex+C-\dfrac{1}{1+xe^x}+C;

(iv) 32(x2+1)2/3+C\tfrac32(x^2+1)^{2/3}+C;

(v) 12log⁡∣e2x−e−2x∣+C\tfrac12\log|e^{2x}-e^{-2x}|+C;

(vi) −18x+78log⁡∣4ex+5e−x∣+C-\tfrac18 x+\tfrac78\log|4e^x+5e^{-x}|+C;

(vii) log⁡∣x2−3x−18∣+C\log|x^2-3x-18|+C;

(viii) −11+log⁡x+C-\dfrac{1}{1+\log x}+C;

(ix) 1alog⁡∣ax+xa∣+C\dfrac1a\log|a^x+x^a|+C.

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