Evaluate: _2^4 (x^2+x ) 2 x+1 d x - Vidyadip

MCQ

Evaluate: $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$

A
$57-5 \sqrt{5}$
B
$\frac{57-\sqrt{5}}{5}$
C
$\frac{57+5 \sqrt{5}}{5}$
✓
$\frac{57-5 \sqrt{5}}{5}$

✓

Answer

Correct option: D.

$\frac{57-5 \sqrt{5}}{5}$

(d) : We have, $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$
Integrating by parts, we get
$
\begin{array}{l}
\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x=\left[\left(x^2+x\right) \cdot \sqrt{2 x+1}\right]_2^4-\int_2^4(2 x+1) \cdot \sqrt{2 x+1} d x \\
=(60-6 \sqrt{5})-\int_2^4(2 x+1)^{3 / 2} d x \\
=(60-6 \sqrt{5})-\frac{1}{5} \cdot\left[(2 x+1)^{5 / 2}\right]_2^4 \\
=(60-6 \sqrt{5})-\left(\frac{243}{5}-5 \sqrt{5}\right)=\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)
\end{array}
$

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 Integrals→Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let f : R → R be defined as $\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}.$ Then, find f(-1) + f(2) + f(4):

9
14
5
None of these.

$\sin [{\cot ^{ - 1}}(\cos {\tan ^{ - 1}}x)] =$

$\int_0^\infty {{e^{ - 2x}}(\sin 2x + \cos 2x)\,dx = } $

If $y=\sin \left(2 \sin ^{-1} x\right)$, then $\left(1-x^2\right) y_2$ is equal to

If A and B are two independent events such that P(A) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then P(A|B) - P(B|A) =

The solution of $y{e^{ - x/y}}dx - (x{e^{ - x/y}} + {y^3})dy = 0$ is

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+2|\vec{b}|^{2}, \vec{a} \cdot \vec{b}=3 \quad$ and $\quad|\vec{a} \times \vec{b}|^{2}=75$.Then $|\vec{a}|^{2}$ is equal to $.......$

Choose the correct answer from the given four options.

Which of the following is the principal value branch of cos^-1x?

$\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
$(0,\pi)$
$[0,\pi]$
$(0,\pi)-\Big\{\frac{\pi}{2}\Big\}$

The solution of $\frac{{dy}}{{dx}} + \frac{y}{3} = 1$ is

$2{x^3} - 6x + 5$ is an increasing function if