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Miscellaneous · Q2

Q.Evaluate the following:

(i) ∫23x3+1x(x−1) dx\int_2^3 \frac{x^3+1}{x(x-1)}\,dx
(ii) ∫1/31(x−x3)1/3x4 dx\int_{1/3}^{1} \frac{(x-x^3)^{1/3}}{x^4}\,dx
(iii) ∫01log⁡(1x−1)dx\int_0^1 \log\left(\frac{1}{x}-1\right)dx
(iv) ∫02x22−x dx\int_0^2 x^2\sqrt{2-x}\,dx
(v) ∫−1111+ex3 dx\int_{-1}^{1} \frac{1}{1+e^{x^3}}\,dx
(vi) ∫−11∣x∣−x dx\int_{-1}^{1} \sqrt{|x|-x}\,dx
Sikkim CbseNCERTSubjective· 5mImportance★★★★★
37% · 22/59 Questions
✓ Free question

Six definite integrals evaluated using division/partial fractions, substitution, and the symmetry properties ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx and ∫−aaf(x) dx=∫0a[f(x)+f(−x)] dx\int_{-a}^a f(x)\,dx=\int_0^a[f(x)+f(-x)]\,dx.

∫abf(x) dx=F(b)−F(a);∫−aaf(x) dx=∫0a(f(x)+f(−x)) dx;∫0af(x) dx=∫0af(a−x) dx.\displaystyle\int_a^b f(x)\,dx=F(b)-F(a);\quad \int_{-a}^{a}f(x)\,dx=\int_0^a\big(f(x)+f(-x)\big)\,dx;\quad \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx.

(i) ∫23x3+1x(x−1) dx\displaystyle\int_2^3 \frac{x^3+1}{x(x-1)}\,dx

  1. Divide: x3+1=(x2−x)(x+1)+(x+1)x^3+1=(x^2-x)(x+1)+(x+1), so x3+1x(x−1)=(x+1)+x+1x(x−1)\dfrac{x^3+1}{x(x-1)}=(x+1)+\dfrac{x+1}{x(x-1)}.
  2. Partial fractions x+1x(x−1)=Ax+Bx−1\dfrac{x+1}{x(x-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}: x+1=A(x−1)+Bxx+1=A(x-1)+Bx. At x=0, A=−1x=0,\ A=-1; at x=1, B=2x=1,\ B=2.
  3. Antiderivative: ∫[(x+1)−1x+2x−1]dx=x22+x−log⁡∣x∣+2log⁡∣x−1∣.\displaystyle\int\Big[(x+1)-\frac1x+\frac2{x-1}\Big]dx=\frac{x^2}{2}+x-\log|x|+2\log|x-1|.
  4. At x=3x=3: 92+3−ln⁡3+2ln⁡2=152−ln⁡3+2ln⁡2\frac92+3-\ln3+2\ln2=\frac{15}{2}-\ln3+2\ln2. At x=2x=2: 2+2−ln⁡2+0=4−ln⁡22+2-\ln2+0=4-\ln2.
  5. Difference =152−4−ln⁡3+3ln⁡2=72+log⁡83≈3.5+0.981=4.48.=\frac{15}{2}-4-\ln3+3\ln2=\frac72+\log\frac{8}{3}\approx 3.5+0.981=4.48.

(ii) ∫1/31(x−x3)1/3x4 dx\displaystyle\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx

  1. Factor x−x3=x3(1x2−1)x-x^3=x^3\big(\tfrac1{x^2}-1\big), so (x−x3)1/3=x(1x2−1)1/3(x-x^3)^{1/3}=x\big(\tfrac1{x^2}-1\big)^{1/3} and the integrand =(1x2−1)1/3x3=\dfrac{(\frac1{x^2}-1)^{1/3}}{x^3}.
  2. Put u=1x2−1⇒du=−2x3dx⇒dxx3=−12 duu=\dfrac1{x^2}-1\Rightarrow du=-\dfrac{2}{x^3}dx\Rightarrow \dfrac{dx}{x^3}=-\tfrac12\,du.
  3. Limits: x=13⇒u=9−1=8x=\tfrac13\Rightarrow u=9-1=8; x=1⇒u=0x=1\Rightarrow u=0.
  4. ∫80u1/3(−12)du=12∫08u1/3du=12⋅34[u4/3]08=38⋅84/3=38⋅16=6.\displaystyle\int_{8}^{0}u^{1/3}\big(-\tfrac12\big)du=\tfrac12\int_0^8 u^{1/3}du=\tfrac12\cdot\frac{3}{4}\big[u^{4/3}\big]_0^8=\frac{3}{8}\cdot 8^{4/3}=\frac38\cdot16=6.

(iii) ∫01log⁡(1x−1)dx=∫01log⁡1−xx dx\displaystyle\int_0^1\log\Big(\frac1x-1\Big)dx=\int_0^1\log\frac{1-x}{x}\,dx

  1. Let I=∫01log⁡1−xxdxI=\int_0^1\log\frac{1-x}{x}dx. Replace x→1−xx\to 1-x: I=∫01log⁡x1−xdx=−II=\int_0^1\log\frac{x}{1-x}dx=-I.
  2. Hence 2I=0⇒I=0.2I=0\Rightarrow I=0.

(iv) ∫02x22−x dx\displaystyle\int_0^2 x^2\sqrt{2-x}\,dx

  1. Put t=2−x⇒x=2−t, dx=−dtt=2-x\Rightarrow x=2-t,\ dx=-dt; limits x=0→t=2, x=2→t=0x{=}0\to t{=}2,\ x{=}2\to t{=}0.
  2. x2=(2−t)2=4−4t+t2x^2=(2-t)^2=4-4t+t^2, so I=∫02(4−4t+t2)t1/2dt=∫02(4t1/2−4t3/2+t5/2)dtI=\int_0^2(4-4t+t^2)t^{1/2}dt=\int_0^2\big(4t^{1/2}-4t^{3/2}+t^{5/2}\big)dt.
  3. =[83t3/2−85t5/2+27t7/2]02=\Big[\tfrac83 t^{3/2}-\tfrac85 t^{5/2}+\tfrac27 t^{7/2}\Big]_0^2 with 23/2=22, 25/2=42, 27/2=822^{3/2}=2\sqrt2,\ 2^{5/2}=4\sqrt2,\ 2^{7/2}=8\sqrt2.
  4. =1623−3225+1627=(560−672+240)2105=1282105≈1.724.=\tfrac{16\sqrt2}{3}-\tfrac{32\sqrt2}{5}+\tfrac{16\sqrt2}{7}=\dfrac{(560-672+240)\sqrt2}{105}=\dfrac{128\sqrt2}{105}\approx 1.724.

(v) ∫−11dx1+ex3\displaystyle\int_{-1}^{1}\frac{dx}{1+e^{x^3}}

  1. Let f(x)=11+ex3f(x)=\dfrac1{1+e^{x^3}}. Then f(−x)=11+e−x3=ex31+ex3f(-x)=\dfrac1{1+e^{-x^3}}=\dfrac{e^{x^3}}{1+e^{x^3}}, so f(x)+f(−x)=1f(x)+f(-x)=1.
  2. I=∫01(f(x)+f(−x))dx=∫011 dx=1.\displaystyle I=\int_0^1\big(f(x)+f(-x)\big)dx=\int_0^1 1\,dx=1.

(vi) ∫−11∣x∣−x dx\displaystyle\int_{-1}^{1}\sqrt{|x|-x}\,dx

  1. For x≥0: ∣x∣−x=0x\ge 0:\ |x|-x=0; for x<0: ∣x∣−x=−x−x=−2xx<0:\ |x|-x=-x-x=-2x.
  2. I=∫−10−2x dx+∫010 dx=2∫−10−x dx.\displaystyle I=\int_{-1}^{0}\sqrt{-2x}\,dx+\int_0^1 0\,dx=\sqrt2\int_{-1}^{0}\sqrt{-x}\,dx.
  3. Put t=−xt=-x: =2∫01t dt=2⋅23=223≈0.943.=\sqrt2\int_0^1\sqrt t\,dt=\sqrt2\cdot\tfrac23=\dfrac{2\sqrt2}{3}\approx 0.943.
✓Final answer

  1. 72+log⁡83≈4.48\dfrac{7}{2}+\log\dfrac{8}{3}\approx 4.48;
  2. 66;
  3. 00;
  4. 1282105≈1.724\dfrac{128\sqrt2}{105}\approx 1.724;
  5. 11;
  6. 223≈0.943\dfrac{2\sqrt2}{3}\approx 0.943.

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