The integral $16 \int_{1}^{2} \frac{\mathrm{dx}}{\mathrm{x}^{3}\left(\mathrm{x}^{2}+2\right)^{2}}$ is equal to

(D) $\frac{11}{6}-\log _{e} 4$

(A) $\frac{11}{6}+\log _{e} 4$

(B) $\frac{11}{12}+\log _{e} 4$

(C) $\frac{11}{12}-\log _{e} 4$

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Indefinite Integrals

The integral

16 \int_{1}^{2} \frac{dx}{x ^{3} ( x ^{2} + 2 ) ^{2}}

is equal to

Answer: (D)

Substituting

u = x^{2}

turns the integral into

16 \int_{1}^{2} \frac{d x}{x ^{3} ( x ^{2} + 2 ) ^{2}} = 8 \int_{1}^{4} \frac{d u}{u ^{2} ( u + 2 ) ^{2}}

. Partial fractions give

\frac{1}{u ^{2} ( u + 2 ) ^{2}} = - \frac{1}{4 u} + \frac{1}{4 u ^{2}} + \frac{1}{4 ( u + 2 )} + \frac{1}{4 ( u + 2 ) ^{2}}

, whose integral from

1

4

\frac{11}{48} - \frac{1}{4} ln 2

. Multiplying by

8

gives

\frac{11}{6} - 2 ln 2 = \frac{11}{6} - ln 4

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option