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Miscellaneous · Q1

Q.Integrate the following:

(i) x3ex2x^3e^{x^2}
(ii) ∫(x4−x)1/4x5 dx\int \frac{(x^4-x)^{1/4}}{x^5}\,dx
(iii) ∫x3+xx4−9 dx\int \frac{x^3+x}{x^4-9}\,dx
(iv) ∫2x4x−1 dx\int \frac{2^x}{\sqrt{4^x-1}}\,dx
(v) ∫1(ex+1)2 dx\int \frac{1}{(e^x+1)^2}\,dx
(vi) ∫(1+x)log⁡x dx\int (1+x)\log x\,dx
Yanam CbseNCERTSubjective· 5mImportance★★★★★
36% · 21/59 Questions
✓ Free question

Six indefinite integrals worked by substitution, factoring out a power, partial fractions and integration by parts.

By parts: ∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du. ∫dtt2−1=log⁡∣t+t2−1∣\int\frac{dt}{\sqrt{t^2-1}}=\log|t+\sqrt{t^2-1}|. ∫dvv2−a2=12alog⁡∣v−av+a∣\int\frac{dv}{v^2-a^2}=\frac{1}{2a}\log\left|\frac{v-a}{v+a}\right|.

(i) ∫x3ex2 dx\int x^3e^{x^2}\,dx. Put t=x2, dt=2x dxt=x^2,\ dt=2x\,dx, so x3 dx=t⋅dt2x^3\,dx=t\cdot\tfrac{dt}{2}:

=12∫t et dt=12(tet−et)=12ex2(x2−1)+C\displaystyle=\tfrac12\int t\,e^{t}\,dt=\tfrac12\big(t e^{t}-e^{t}\big)=\tfrac12 e^{x^2}(x^2-1)+C.

(ii) ∫(x4−x)1/4x5 dx\int\frac{(x^4-x)^{1/4}}{x^5}\,dx. Write x4−x=x4 ⁣(1−x−3)x^4-x=x^4\!\left(1-x^{-3}\right), so (x4−x)1/4=x(1−x−3)1/4(x^4-x)^{1/4}=x\left(1-x^{-3}\right)^{1/4} and the integrand =(1−x−3)1/4x4=\dfrac{\left(1-x^{-3}\right)^{1/4}}{x^4}. Put t=1−x−3, dt=3x−4 dxt=1-x^{-3},\ dt=3x^{-4}\,dx:

=13∫t1/4 dt=13⋅45t5/4=415(1−1x3)5/4+C\displaystyle=\tfrac13\int t^{1/4}\,dt=\tfrac13\cdot\tfrac{4}{5}t^{5/4}=\tfrac{4}{15}\left(1-\tfrac{1}{x^3}\right)^{5/4}+C.

(iii) ∫x3+xx4−9 dx=∫x3x4−9 dx+∫xx4−9 dx\int\frac{x^3+x}{x^4-9}\,dx=\int\frac{x^3}{x^4-9}\,dx+\int\frac{x}{x^4-9}\,dx.

First: u=x4−9, du=4x3 dx⇒14log⁡∣x4−9∣u=x^4-9,\ du=4x^3\,dx\Rightarrow\tfrac14\log|x^4-9|.

Second: v=x2, dv=2x dx⇒12∫dvv2−9=12⋅16log⁡∣v−3v+3∣=112log⁡∣x2−3x2+3∣v=x^2,\ dv=2x\,dx\Rightarrow\tfrac12\int\frac{dv}{v^2-9}=\tfrac12\cdot\tfrac{1}{6}\log\left|\tfrac{v-3}{v+3}\right|=\tfrac{1}{12}\log\left|\tfrac{x^2-3}{x^2+3}\right|.

Sum =14log⁡∣x4−9∣+112log⁡∣x2−3x2+3∣+C=\tfrac14\log|x^4-9|+\tfrac{1}{12}\log\left|\dfrac{x^2-3}{x^2+3}\right|+C.

(iv) ∫2x4x−1 dx\int\frac{2^x}{\sqrt{4^x-1}}\,dx. Put t=2x, dt=2xln⁡2 dxt=2^x,\ dt=2^x\ln2\,dx, and 4x=t24^x=t^2:

=1ln⁡2∫dtt2−1=1ln⁡2log⁡∣t+t2−1∣=1ln⁡2log⁡∣2x+4x−1∣+C\displaystyle=\tfrac{1}{\ln2}\int\frac{dt}{\sqrt{t^2-1}}=\tfrac{1}{\ln2}\log\left|t+\sqrt{t^2-1}\right|=\tfrac{1}{\ln2}\log\left|2^x+\sqrt{4^x-1}\right|+C.

(v) ∫dx(ex+1)2\int\frac{dx}{(e^x+1)^2}. Put t=ex, dx=dttt=e^x,\ dx=\tfrac{dt}{t}: ∫dtt(t+1)2\int\frac{dt}{t(t+1)^2}. Partial fractions 1t(t+1)2=1t−1t+1−1(t+1)2\frac{1}{t(t+1)^2}=\frac{1}{t}-\frac{1}{t+1}-\frac{1}{(t+1)^2}:

=log⁡t−log⁡(t+1)+1t+1=log⁡exex+1+1ex+1=x−log⁡(ex+1)+1ex+1+C\displaystyle=\log t-\log(t+1)+\frac{1}{t+1}=\log\frac{e^x}{e^x+1}+\frac{1}{e^x+1}=x-\log(e^x+1)+\frac{1}{e^x+1}+C.

(vi) ∫(1+x)log⁡x dx\int(1+x)\log x\,dx. By parts with u=log⁡x, dv=(1+x) dx, v=x+x22u=\log x,\ dv=(1+x)\,dx,\ v=x+\tfrac{x^2}{2}:

=(x+x22)log⁡x−∫(x+x22)1x dx=(x+x22)log⁡x−∫(1+x2)dx=(x+x22)log⁡x−x−x24+C\displaystyle=\Big(x+\tfrac{x^2}{2}\Big)\log x-\int\Big(x+\tfrac{x^2}{2}\Big)\tfrac{1}{x}\,dx=\Big(x+\tfrac{x^2}{2}\Big)\log x-\int\Big(1+\tfrac{x}{2}\Big)dx=\Big(x+\tfrac{x^2}{2}\Big)\log x-x-\tfrac{x^2}{4}+C.

✓Final answer

  1. 12ex2(x2−1)+C\tfrac12 e^{x^2}(x^2-1)+C;
  2. 415(1−1x3)5/4+C\tfrac{4}{15}\left(1-\tfrac{1}{x^3}\right)^{5/4}+C;
  3. 14log⁡∣x4−9∣+112log⁡∣x2−3x2+3∣+C\tfrac14\log|x^4-9|+\tfrac{1}{12}\log\left|\tfrac{x^2-3}{x^2+3}\right|+C;
  4. 1ln⁡2log⁡∣2x+4x−1∣+C\tfrac{1}{\ln2}\log\left|2^x+\sqrt{4^x-1}\right|+C;
  5. x−log⁡(ex+1)+1ex+1+Cx-\log(e^x+1)+\tfrac{1}{e^x+1}+C;
  6. (x+x22)log⁡x−x−x24+C\left(x+\tfrac{x^2}{2}\right)\log x-x-\tfrac{x^2}{4}+C.

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