The integral 16 _1^2 d x x^3 (x^2+2 )^2 is equal to - Vidyadip

MCQ

The integral $16 \int \limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$ is equal to

A
$\frac{11}{6}+\log _e 4$
B
$\frac{11}{12}+\log _e 4$
C
$\frac{11}{12}-\log _{ e } 4$
✓
$\frac{11}{6}-\log _e 4$

✓

Answer

Correct option: D.

$\frac{11}{6}-\log _e 4$

d
$I=16 \int \limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$

$=16 \int \limits_1^2 \frac{d x}{x^3 x^4\left(1+\frac{2}{x^2}\right)^2}$

Let, $1+\frac{2}{x^2}=t \Rightarrow \frac{-4}{x^3} d x=d t$

$I=-4 \int \limits_3^{\frac{3}{2}} \frac{d t}{\left(\frac{2}{t-1}\right)^2 t^2}$

$I=-4 \int \limits_3^{\frac{3}{2}}\left(\frac{t-1}{2}\right)^2 \frac{d t}{t^2}$

$I=-\frac{4}{4} \int \limits_3^{\frac{3}{2}}\left(1-\frac{2}{t}+\frac{1}{t^2}\right) d t$

$I=-1\left[t-2 \ell n|t|-\frac{1}{t}\right]_3^{\frac{3}{2}}$

$I=-1\left[\left(\frac{3}{2}-2 \ell n \frac{3}{2}-\frac{2}{3}\right)-\left(3-2 \ell n 3-\frac{1}{3}\right)\right]$

$I=-1\left[2 \ell n 2-\frac{11}{6}\right]$

$I=\frac{11}{6}-\ell n 4$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $A = \left( {\begin{array}{*{20}{c}}
{\cos \,\alpha }&{ - \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ such that ${A^{32}} = \left( {\begin{array}{*{20}{c}}
0&{ - 1}\\
1&0
\end{array}} \right)$. Then a value of $\alpha $ is

$\int_{}^{} {\frac{{\tan x}}{{\sec x + \tan x}}\;dx = } $

If $f(x)=x^3-6 x^2+9 x+3$ be a decreasing function, then $x$ lies in

$\int \limits_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$ is equal to

Let $f :\left[\frac{1}{2}, 1\right] \rightarrow R$ (the set of all real numbers) be a positive, non-constant and differentiable function such that $f^{\prime}(x)<2 f(x)$ and $f\left(\frac{1}{2}\right)=1$. Then the value of $\int_{1 / 2}^1 f(x) d x$ lies in the interval

$\tan \left[ {{{\cos }^{ - 1}}\frac{4}{5} + {{\tan }^{ - 1}}\frac{2}{3}} \right] =$

The maximum volume (in $cu.m$) of the right circular cone having slant height $3\, m$ is

If $\int {} e^u$ . $\sin 2x dx$ can be found in terms of known functions of $x$ then $u$ can be :

$\int\limits^\frac{\pi}{2}_0\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$ equals to:

$f : R \rightarrow R$ is defined by $\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$ is: