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3.3 · Q1

Q.Integrate the following functions:

(i) xe2x+3xe^{2x+3}
(ii) xlog⁡(x2+1)x\log(x^2+1)
(iii) x2exx^2e^x
(iv) xlog⁡xx\log x
(v) xlog⁡2xx\log 2x
(vi) x2log⁡xx^2\log x
(vii) (x2+1)log⁡x(x^2+1)\log x
(viii) x(log⁡x)2x(\log x)^2
Sikkim CbseNCERTSubjective· 5mImportance★★★★★
46% · 27/59 Questions
✓ Free question

Each is integration by parts, ∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du, taking log⁡\log or the power as uu.

By parts: ∫u dv=uv−∫v du\displaystyle\int u\,dv=uv-\int v\,du (choose uu by ILATE: Inverse, Log, Algebraic, Trig, Exponential).

(i) ∫xe2x+3dx\displaystyle\int xe^{2x+3}dx

  1. u=x, dv=e2x+3dx⇒v=12e2x+3u=x,\ dv=e^{2x+3}dx\Rightarrow v=\tfrac12e^{2x+3}.
  2. =x2e2x+3−∫12e2x+3dx=x2e2x+3−14e2x+3+C=e2x+34(2x−1)+C=\tfrac{x}{2}e^{2x+3}-\int\tfrac12e^{2x+3}dx=\tfrac{x}{2}e^{2x+3}-\tfrac14e^{2x+3}+C=\tfrac{e^{2x+3}}{4}(2x-1)+C.

(ii) ∫xlog⁡(x2+1) dx\displaystyle\int x\log(x^2+1)\,dx

  1. u=log⁡(x2+1), dv=x dx⇒v=x22u=\log(x^2+1),\ dv=x\,dx\Rightarrow v=\tfrac{x^2}{2}; du=2xx2+1dxdu=\frac{2x}{x^2+1}dx.
  2. =x22log⁡(x2+1)−∫x22⋅2xx2+1dx=x22log⁡(x2+1)−∫x3x2+1dx=\tfrac{x^2}{2}\log(x^2+1)-\int\tfrac{x^2}{2}\cdot\tfrac{2x}{x^2+1}dx=\tfrac{x^2}{2}\log(x^2+1)-\int\tfrac{x^3}{x^2+1}dx.
  3. x3x2+1=x−xx2+1⇒∫=x22−12log⁡(x2+1)\tfrac{x^3}{x^2+1}=x-\tfrac{x}{x^2+1}\Rightarrow \int=\tfrac{x^2}{2}-\tfrac12\log(x^2+1).
  4. =x22log⁡(x2+1)−x22+12log⁡(x2+1)+C=x2+12log⁡(x2+1)−x22+C=\tfrac{x^2}{2}\log(x^2+1)-\tfrac{x^2}{2}+\tfrac12\log(x^2+1)+C=\tfrac{x^2+1}{2}\log(x^2+1)-\tfrac{x^2}{2}+C.

(iii) ∫x2exdx\displaystyle\int x^2e^x dx

  1. =x2ex−2∫xexdx=x^2e^x-2\int xe^x dx; and ∫xexdx=xex−ex\int xe^x dx=xe^x-e^x.
  2. =x2ex−2(xex−ex)=ex(x2−2x+2)+C=x^2e^x-2(xe^x-e^x)=e^x(x^2-2x+2)+C.

(iv) ∫xlog⁡x dx\displaystyle\int x\log x\,dx

  1. u=log⁡x, v=x22u=\log x,\ v=\tfrac{x^2}{2}: =x22log⁡x−∫x22⋅1xdx=x22log⁡x−x24+C=\tfrac{x^2}{2}\log x-\int\tfrac{x^2}{2}\cdot\tfrac1x dx=\tfrac{x^2}{2}\log x-\tfrac{x^2}{4}+C.

(v) ∫xlog⁡2x dx\displaystyle\int x\log 2x\,dx

  1. u=log⁡2x (du=1xdx), v=x22u=\log 2x\ (du=\tfrac1x dx),\ v=\tfrac{x^2}{2}: =x22log⁡2x−∫x22⋅1xdx=x22log⁡2x−x24+C=\tfrac{x^2}{2}\log 2x-\int\tfrac{x^2}{2}\cdot\tfrac1x dx=\tfrac{x^2}{2}\log 2x-\tfrac{x^2}{4}+C.

(vi) ∫x2log⁡x dx\displaystyle\int x^2\log x\,dx

  1. u=log⁡x, v=x33u=\log x,\ v=\tfrac{x^3}{3}: =x33log⁡x−∫x33⋅1xdx=x33log⁡x−x39+C=\tfrac{x^3}{3}\log x-\int\tfrac{x^3}{3}\cdot\tfrac1x dx=\tfrac{x^3}{3}\log x-\tfrac{x^3}{9}+C.

(vii) ∫(x2+1)log⁡x dx\displaystyle\int (x^2+1)\log x\,dx

  1. u=log⁡x, v=x33+xu=\log x,\ v=\tfrac{x^3}{3}+x: =(x33+x)log⁡x−∫(x33+x)1xdx=\Big(\tfrac{x^3}{3}+x\Big)\log x-\int\Big(\tfrac{x^3}{3}+x\Big)\tfrac1x dx.
  2. =(x33+x)log⁡x−∫(x23+1)dx=(x33+x)log⁡x−x39−x+C=\Big(\tfrac{x^3}{3}+x\Big)\log x-\int\Big(\tfrac{x^2}{3}+1\Big)dx=\Big(\tfrac{x^3}{3}+x\Big)\log x-\tfrac{x^3}{9}-x+C.

(viii) ∫x(log⁡x)2dx\displaystyle\int x(\log x)^2 dx

  1. u=(log⁡x)2, v=x22u=(\log x)^2,\ v=\tfrac{x^2}{2}: =x22(log⁡x)2−∫x22⋅2log⁡x⋅1xdx=x22(log⁡x)2−∫xlog⁡x dx=\tfrac{x^2}{2}(\log x)^2-\int\tfrac{x^2}{2}\cdot 2\log x\cdot\tfrac1x dx=\tfrac{x^2}{2}(\log x)^2-\int x\log x\,dx.
  2. Using (iv), ∫xlog⁡x dx=x22log⁡x−x24\int x\log x\,dx=\tfrac{x^2}{2}\log x-\tfrac{x^2}{4}.
  3. =x22(log⁡x)2−x22log⁡x+x24+C=\tfrac{x^2}{2}(\log x)^2-\tfrac{x^2}{2}\log x+\tfrac{x^2}{4}+C.
✓Final answer

(i) e2x+34(2x−1)+C\tfrac{e^{2x+3}}{4}(2x-1)+C (ii) x2+12log⁡(x2+1)−x22+C\tfrac{x^2+1}{2}\log(x^2+1)-\tfrac{x^2}{2}+C (iii) ex(x2−2x+2)+Ce^x(x^2-2x+2)+C (iv) x22log⁡x−x24+C\tfrac{x^2}{2}\log x-\tfrac{x^2}{4}+C (v) x22log⁡2x−x24+C\tfrac{x^2}{2}\log 2x-\tfrac{x^2}{4}+C (vi) x33log⁡x−x39+C\tfrac{x^3}{3}\log x-\tfrac{x^3}{9}+C (vii) (x33+x)log⁡x−x39−x+C\big(\tfrac{x^3}{3}+x\big)\log x-\tfrac{x^3}{9}-x+C (viii) x22(log⁡x)2−x22log⁡x+x24+C\tfrac{x^2}{2}(\log x)^2-\tfrac{x^2}{2}\log x+\tfrac{x^2}{4}+C

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