Skip to content
3.2 · Q1

Q.Integrate the following expressions:

(i) x+1(x+2)(x+4)\frac{x+1}{(x+2)(x+4)}
(ii) x(x2+1)(x2+2)\frac{x}{(x^2+1)(x^2+2)}
(iii) 1e2x−1\frac{1}{e^{2x}-1}
(iv) 1x((log⁡x)2−3log⁡x+2)\frac{1}{x((\log x)^2-3\log x+2)}
(v) 3x−2(x−2)2(x+2)\frac{3x-2}{(x-2)^2(x+2)}
(vi) 1e2x+ex\frac{1}{e^{2x}+e^x}
(vii) 5x+4(x2−1)(x+2)\frac{5x+4}{(x^2-1)(x+2)}
(viii) x(x−1)2(x+2)\frac{x}{(x-1)^2(x+2)}
(ix) 1x(x4−1)\frac{1}{x(x^4-1)}
(x) 1x(xn+1)\frac{1}{x(x^n+1)}
(xi) 1−xx(1−2x)\frac{1-x}{x(1-2x)}
Delhi CbseNCERTSubjective· 5mImportance★★★★★
44% · 26/59 Questions
✓ Free question

Each integrand is split by partial fractions (or a suitable substitution) and integrated term-by-term.

Partial fractions + standard forms: ∫dxx−a=log⁡∣x−a∣\int\frac{dx}{x-a}=\log|x-a|, ∫dx(x−a)2=−1x−a\int\frac{dx}{(x-a)^2}=-\frac{1}{x-a}. For a repeated/irreducible factor, split first; for a substitution set t=t= inner function so dt=dt= its derivative ×dx\times dx.

(i) ∫x+1(x+2)(x+4)dx\displaystyle\int\frac{x+1}{(x+2)(x+4)}dx

  1. x+1(x+2)(x+4)=Ax+2+Bx+4⇒x+1=A(x+4)+B(x+2)\frac{x+1}{(x+2)(x+4)}=\frac{A}{x+2}+\frac{B}{x+4}\Rightarrow x+1=A(x+4)+B(x+2).
  2. x=−2: −1=2A⇒A=−12x=-2:\,-1=2A\Rightarrow A=-\tfrac12; x=−4: −3=−2B⇒B=32x=-4:\,-3=-2B\Rightarrow B=\tfrac32.
  3. ∫=−12log⁡∣x+2∣+32log⁡∣x+4∣+C\int=-\tfrac12\log|x+2|+\tfrac32\log|x+4|+C.

(ii) ∫x(x2+1)(x2+2)dx\displaystyle\int\frac{x}{(x^2+1)(x^2+2)}dx

  1. Put t=x2, dt=2x dxt=x^2,\ dt=2x\,dx: integral =12∫dt(t+1)(t+2)=\tfrac12\int\frac{dt}{(t+1)(t+2)}.
  2. 1(t+1)(t+2)=1t+1−1t+2\frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}.
  3. =12[log⁡∣t+1∣−log⁡∣t+2∣]=12log⁡x2+1x2+2+C=\tfrac12\big[\log|t+1|-\log|t+2|\big]=\tfrac12\log\dfrac{x^2+1}{x^2+2}+C.

(iii) ∫dxe2x−1\displaystyle\int\frac{dx}{e^{2x}-1}

  1. Write 1e2x−1=e−2x1−e−2x\frac{1}{e^{2x}-1}=\frac{e^{-2x}}{1-e^{-2x}}.
  2. Put t=1−e−2x, dt=2e−2xdxt=1-e^{-2x},\ dt=2e^{-2x}dx: integral =12∫dtt=\tfrac12\int\frac{dt}{t}.
  3. =12log⁡∣1−e−2x∣+C=12log⁡(e2x−1)−x+C=\tfrac12\log|1-e^{-2x}|+C=\tfrac12\log(e^{2x}-1)-x+C.

(iv) ∫dxx((log⁡x)2−3log⁡x+2)\displaystyle\int\frac{dx}{x\big((\log x)^2-3\log x+2\big)}

  1. Put t=log⁡x, dt=dxxt=\log x,\ dt=\frac{dx}{x}: integral =∫dtt2−3t+2=∫dt(t−1)(t−2)=\int\frac{dt}{t^2-3t+2}=\int\frac{dt}{(t-1)(t-2)}.
  2. 1(t−1)(t−2)=1t−2−1t−1\frac{1}{(t-1)(t-2)}=\frac{1}{t-2}-\frac{1}{t-1}.
  3. =log⁡∣log⁡x−2log⁡x−1∣+C=\log\Big|\dfrac{\log x-2}{\log x-1}\Big|+C.

(v) ∫3x−2(x−2)2(x+2)dx\displaystyle\int\frac{3x-2}{(x-2)^2(x+2)}dx

  1. 3x−2(x−2)2(x+2)=Ax−2+B(x−2)2+Cx+2\frac{3x-2}{(x-2)^2(x+2)}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+2}; 3x−2=A(x−2)(x+2)+B(x+2)+C(x−2)23x-2=A(x-2)(x+2)+B(x+2)+C(x-2)^2.
  2. x=2: 4=4B⇒B=1x=2:\,4=4B\Rightarrow B=1; x=−2: −8=16C⇒C=−12x=-2:\,-8=16C\Rightarrow C=-\tfrac12; x2x^2-coeff 0=A+C⇒A=120=A+C\Rightarrow A=\tfrac12.
  3. ∫=12log⁡∣x−2∣−1x−2−12log⁡∣x+2∣+C=12log⁡∣x−2x+2∣−1x−2+C\int=\tfrac12\log|x-2|-\dfrac{1}{x-2}-\tfrac12\log|x+2|+C=\tfrac12\log\Big|\dfrac{x-2}{x+2}\Big|-\dfrac{1}{x-2}+C.

(vi) ∫dxe2x+ex\displaystyle\int\frac{dx}{e^{2x}+e^{x}}

  1. =∫dxex(ex+1)=\int\frac{dx}{e^x(e^x+1)}. Put t=ex, dx=dttt=e^x,\ dx=\frac{dt}{t}: integral =∫dtt2(t+1)=\int\frac{dt}{t^2(t+1)}.
  2. 1t2(t+1)=−1t+1t2+1t+1\frac{1}{t^2(t+1)}=-\frac{1}{t}+\frac{1}{t^2}+\frac{1}{t+1}.
  3. =−log⁡∣t∣−1t+log⁡∣t+1∣=log⁡(ex+1)−x−e−x+C=-\log|t|-\tfrac1t+\log|t+1|=\log(e^x+1)-x-e^{-x}+C.

(vii) ∫5x+4(x2−1)(x+2)dx=∫5x+4(x−1)(x+1)(x+2)dx\displaystyle\int\frac{5x+4}{(x^2-1)(x+2)}dx=\int\frac{5x+4}{(x-1)(x+1)(x+2)}dx

  1. Ax−1+Bx+1+Cx+2\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}; x=1: 9=6A⇒A=32x=1:\,9=6A\Rightarrow A=\tfrac32; x=−1: −1=−2B⇒B=12x=-1:\,-1=-2B\Rightarrow B=\tfrac12; x=−2: −6=3C⇒C=−2x=-2:\,-6=3C\Rightarrow C=-2.
  2. ∫=32log⁡∣x−1∣+12log⁡∣x+1∣−2log⁡∣x+2∣+C\int=\tfrac32\log|x-1|+\tfrac12\log|x+1|-2\log|x+2|+C.

(viii) ∫x(x−1)2(x+2)dx\displaystyle\int\frac{x}{(x-1)^2(x+2)}dx

  1. Ax−1+B(x−1)2+Cx+2\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}; x=1: 1=3B⇒B=13x=1:\,1=3B\Rightarrow B=\tfrac13; x=−2: −2=9C⇒C=−29x=-2:\,-2=9C\Rightarrow C=-\tfrac29; x2x^2-coeff 0=A+C⇒A=290=A+C\Rightarrow A=\tfrac29.
  2. ∫=29log⁡∣x−1∣−13(x−1)−29log⁡∣x+2∣+C=29log⁡∣x−1x+2∣−13(x−1)+C\int=\tfrac29\log|x-1|-\dfrac{1}{3(x-1)}-\tfrac29\log|x+2|+C=\tfrac29\log\Big|\dfrac{x-1}{x+2}\Big|-\dfrac{1}{3(x-1)}+C.

(ix) ∫dxx(x4−1)\displaystyle\int\frac{dx}{x(x^4-1)}

  1. Multiply top & bottom by x3x^3: ∫x3 dxx4(x4−1)\int\frac{x^3\,dx}{x^4(x^4-1)}. Put u=x4, du=4x3dxu=x^4,\ du=4x^3dx: =14∫duu(u−1)=\tfrac14\int\frac{du}{u(u-1)}.
  2. 1u(u−1)=1u−1−1u\frac{1}{u(u-1)}=\frac{1}{u-1}-\frac1u.
  3. =14log⁡∣x4−1x4∣+C=\tfrac14\log\Big|\dfrac{x^4-1}{x^4}\Big|+C.

(x) ∫dxx(xn+1)\displaystyle\int\frac{dx}{x(x^n+1)}

  1. Multiply by xn−1x^{n-1}: ∫xn−1dxxn(xn+1)\int\frac{x^{n-1}dx}{x^n(x^n+1)}. Put u=xn, du=n xn−1dxu=x^n,\ du=n\,x^{n-1}dx: =1n∫duu(u+1)=\tfrac1n\int\frac{du}{u(u+1)}.
  2. 1u(u+1)=1u−1u+1\frac{1}{u(u+1)}=\frac1u-\frac{1}{u+1}.
  3. =1nlog⁡∣xnxn+1∣+C=\tfrac1n\log\Big|\dfrac{x^n}{x^n+1}\Big|+C.

(xi) ∫1−xx(1−2x)dx\displaystyle\int\frac{1-x}{x(1-2x)}dx

  1. 1−xx(1−2x)=Ax+B1−2x\frac{1-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{1-2x}; 1−x=A(1−2x)+Bx1-x=A(1-2x)+Bx; x=0: A=1x=0:\,A=1; xx-coeff −1=−2A+B⇒B=1-1=-2A+B\Rightarrow B=1.
  2. ∫=log⁡∣x∣−12log⁡∣1−2x∣+C\int=\log|x|-\tfrac12\log|1-2x|+C.
✓Final answer

(i) −12log⁡∣x+2∣+32log⁡∣x+4∣+C-\tfrac12\log|x+2|+\tfrac32\log|x+4|+C (ii) 12log⁡x2+1x2+2+C\tfrac12\log\frac{x^2+1}{x^2+2}+C (iii) 12log⁡(e2x−1)−x+C\tfrac12\log(e^{2x}-1)-x+C (iv) log⁡∣log⁡x−2log⁡x−1∣+C\log\big|\frac{\log x-2}{\log x-1}\big|+C (v) 12log⁡∣x−2x+2∣−1x−2+C\tfrac12\log\big|\frac{x-2}{x+2}\big|-\frac{1}{x-2}+C (vi) log⁡(ex+1)−x−e−x+C\log(e^x+1)-x-e^{-x}+C (vii) 32log⁡∣x−1∣+12log⁡∣x+1∣−2log⁡∣x+2∣+C\tfrac32\log|x-1|+\tfrac12\log|x+1|-2\log|x+2|+C (viii) 29log⁡∣x−1x+2∣−13(x−1)+C\tfrac29\log\big|\frac{x-1}{x+2}\big|-\frac{1}{3(x-1)}+C (ix) 14log⁡∣x4−1x4∣+C\tfrac14\log\big|\frac{x^4-1}{x^4}\big|+C (x) 1nlog⁡∣xnxn+1∣+C\tfrac1n\log\big|\frac{x^n}{x^n+1}\big|+C (xi) log⁡∣x∣−12log⁡∣1−2x∣+C\log|x|-\tfrac12\log|1-2x|+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.