The integral 16\int\limits_1^2 dx \over x^3( x^2 + 2 )^2 is equal to

Definite Integration PYQ — JEE Main 2023 · 25 Jan · Shift 2 · Q32 with Solution

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Correct answer: D

We need to evaluate $I=16\int_1^2 \frac{dx}{x^3(x^2+2)^2}.$
Let $t=x^2 \implies dt=2x\,dx, \quad dx=\frac{dt}{2x}.$ A cleaner way is to rewrite using $x^2=t$ , so that $x^3=x\cdot x^2=x t.$ Then $\frac{dx}{x^3(x^2+2)^2}=\frac{dx}{x t (t+2)^2}.$ Using $dx=\frac{dt}{2x}$ and $x^2=t$ ,

=\frac{dt}{2x^2 t (t+2)^2} =\frac{dt}{2t^2(t+2)^2}.$$ So $$I=16\int_{1}^{2} \frac{dx}{x^3(x^2+2)^2} =16\int_{t=1}^{4} \frac{dt}{2t^2(t+2)^2} =8\int_1^4 \frac{dt}{t^2(t+2)^2}.$$ 3. Now decompose $$\frac{8}{t^2(t+2)^2}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t+2}+\frac{D}{(t+2)^2}.$$ Multiplying by $t^2(t+2)^2$: $$8=A t(t+2)^2+B(t+2)^2+C t^2(t+2)+Dt^2.$$ Expand: $$8=A(t^3+4t^2+4t)+B(t^2+4t+4)+C(t^3+2t^2)+Dt^2.$$ So $$8=(A+C)t^3+(4A+B+2C+D)t^2+(4A+4B)t+4B.$$ Comparing coefficients with constant $8$: \[ A+C=0, \] \[ 4A+B+2C+D=0, \] \[ 4A+4B=0, \] \[ 4B=8. \] From $4B=8$, we get $B=2$. From $4A+4B=0$, $A=-2$. From $A+C=0$, $C=2$. From $4A+B+2C+D=0$, $$4(-2)+2+2(2)+D=0 \implies -8+2+4+D=0 \implies D=2.$$ Thus $$\frac{8}{t^2(t+2)^2}=-\frac{2}{t}+\frac{2}{t^2}+\frac{2}{t+2}+\frac{2}{(t+2)^2}.$$ 4. Therefore $$I=\int_1^4 \left(-\frac{2}{t}+\frac{2}{t^2}+\frac{2}{t+2}+\frac{2}{(t+2)^2}\right)dt.$$ Integrate termwise: $$\int -\frac{2}{t}\,dt=-2\ln t,$$ $$\int \frac{2}{t^2}\,dt=2\int t^{-2}dt=-\frac{2}{t},$$ $$\int \frac{2}{t+2}\,dt=2\ln(t+2),$$ $$\int \frac{2}{(t+2)^2}\,dt=-\frac{2}{t+2}.$$ Hence $$I=\left[-2\ln t-\frac{2}{t}+2\ln(t+2)-\frac{2}{t+2}\right]_1^4.$$ 5. Evaluate at the limits. At $t=4$: $$-2\ln 4-\frac{2}{4}+2\ln 6-\frac{2}{6} =-2\ln 4+2\ln 6-\frac12-\frac13 =2\ln\frac{6}{4}-\frac{5}{6} =2\ln\frac32-\frac56.$$ At $t=1$: $$-2\ln 1-2+2\ln 3-\frac{2}{3} =0-2+2\ln 3-\frac23 =2\ln 3-\frac83.$$ So $$I=\left(2\ln\frac32-\frac56\right)-\left(2\ln 3-\frac83\right).$$ Simplify the logarithm part: $$2\ln\frac32-2\ln 3=2\ln\frac{(3/2)}{3}=2\ln\frac12=-2\ln 2=-\ln 4.$$ Simplify the constant part: $$-\frac56+\frac83=-\frac56+\frac{16}{6}=\frac{11}{6}.$$ Thus $$I=\frac{11}{6}-\ln 4.$$ 6. Compare with the options: $$\frac{11}{6}-\log_e 4,$$ which is **Option D**.

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View written solutionFree

Correct answer: D

We need to evaluate $I=16\int_1^2 \frac{dx}{x^3(x^2+2)^2}.$
Let $t=x^2 \implies dt=2x\,dx, \quad dx=\frac{dt}{2x}.$ A cleaner way is to rewrite using $x^2=t$ , so that $x^3=x\cdot x^2=x t.$ Then $\frac{dx}{x^3(x^2+2)^2}=\frac{dx}{x t (t+2)^2}.$ Using $dx=\frac{dt}{2x}$ and $x^2=t$ ,

=\frac{dt}{2x^2 t (t+2)^2} =\frac{dt}{2t^2(t+2)^2}.$$ So $$I=16\int_{1}^{2} \frac{dx}{x^3(x^2+2)^2} =16\int_{t=1}^{4} \frac{dt}{2t^2(t+2)^2} =8\int_1^4 \frac{dt}{t^2(t+2)^2}.$$ 3. Now decompose $$\frac{8}{t^2(t+2)^2}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t+2}+\frac{D}{(t+2)^2}.$$ Multiplying by $t^2(t+2)^2$: $$8=A t(t+2)^2+B(t+2)^2+C t^2(t+2)+Dt^2.$$ Expand: $$8=A(t^3+4t^2+4t)+B(t^2+4t+4)+C(t^3+2t^2)+Dt^2.$$ So $$8=(A+C)t^3+(4A+B+2C+D)t^2+(4A+4B)t+4B.$$ Comparing coefficients with constant $8$: \[ A+C=0, \] \[ 4A+B+2C+D=0, \] \[ 4A+4B=0, \] \[ 4B=8. \] From $4B=8$, we get $B=2$. From $4A+4B=0$, $A=-2$. From $A+C=0$, $C=2$. From $4A+B+2C+D=0$, $$4(-2)+2+2(2)+D=0 \implies -8+2+4+D=0 \implies D=2.$$ Thus $$\frac{8}{t^2(t+2)^2}=-\frac{2}{t}+\frac{2}{t^2}+\frac{2}{t+2}+\frac{2}{(t+2)^2}.$$ 4. Therefore $$I=\int_1^4 \left(-\frac{2}{t}+\frac{2}{t^2}+\frac{2}{t+2}+\frac{2}{(t+2)^2}\right)dt.$$ Integrate termwise: $$\int -\frac{2}{t}\,dt=-2\ln t,$$ $$\int \frac{2}{t^2}\,dt=2\int t^{-2}dt=-\frac{2}{t},$$ $$\int \frac{2}{t+2}\,dt=2\ln(t+2),$$ $$\int \frac{2}{(t+2)^2}\,dt=-\frac{2}{t+2}.$$ Hence $$I=\left[-2\ln t-\frac{2}{t}+2\ln(t+2)-\frac{2}{t+2}\right]_1^4.$$ 5. Evaluate at the limits. At $t=4$: $$-2\ln 4-\frac{2}{4}+2\ln 6-\frac{2}{6} =-2\ln 4+2\ln 6-\frac12-\frac13 =2\ln\frac{6}{4}-\frac{5}{6} =2\ln\frac32-\frac56.$$ At $t=1$: $$-2\ln 1-2+2\ln 3-\frac{2}{3} =0-2+2\ln 3-\frac23 =2\ln 3-\frac83.$$ So $$I=\left(2\ln\frac32-\frac56\right)-\left(2\ln 3-\frac83\right).$$ Simplify the logarithm part: $$2\ln\frac32-2\ln 3=2\ln\frac{(3/2)}{3}=2\ln\frac12=-2\ln 2=-\ln 4.$$ Simplify the constant part: $$-\frac56+\frac83=-\frac56+\frac{16}{6}=\frac{11}{6}.$$ Thus $$I=\frac{11}{6}-\ln 4.$$ 6. Compare with the options: $$\frac{11}{6}-\log_e 4,$$ which is **Option D**.

Definite Integration question

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