2x (2x+1)^2 dx - Vidyadip

Question

$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$

✓

Answer

$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$
$=\int\bigg(\frac{2\text{x}+1-1}{(2\text{x}+1)^2}\bigg)\text{dx}$
$=\int\bigg[\frac{2\text{x}+1}{(2\text{x}+1)^2}-\frac{1}{(2\text{x}+1)^2}\bigg]\text{dx}$
$=\int\frac{\text{dx}}{2\text{x}+1}-\int(2\text{x}+1)^{-2}\text{dx}$
$=\frac{\log(2\text{x}+1)}{2}-\bigg[\frac{(2\text{x}+1)^{-2+1}}{2(-2+1)}\bigg]+\text{c}$
$=\frac{\log(2\text{x}+1)}{2}+\frac{(2\text{x}+1)^{-1}}{2}+\text{c}$
$=\frac{\log(2\text{x}+1)}{2}+\frac{1}{2(2\text{x}+1)}+\text{c}$

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 "Solve the Following Question.(3 Marks)" from Indefinite Integrals→Indefinite Integrals — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $A^{-1}=\frac{1}{3}\left[\begin{array}{ccc}1 & 4 & -2 \\ -2 & -5 & 4 \\ 1 & -2 & 1\end{array}\right]$ and $|A|=3$, then (adj. $\left.A\right)=\ldots$.

Differentiate the following functions with respect to x:
$\text{e}^{\tan3\text{x}}$

Let $A$ and $B$ be two independent events such that $P(A)=p_1$ and $P(B)=p_2$. Describe in words the events whose probabilities are:
$1-\left(1-p_1\right)\left(1-p_2\right)$.

Integrate the following w.r.t. x:

$\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}$

Find the length of the perpendicular drow from the point (5, 4, -1) to the line $\vec{\text{r}}=\hat{\text{i}}+\lambda\big(2\hat{\text{i}}+9\hat{\text{j}}+5\hat{\text{k}}\big).$

Prove that the function $f$ given by $f(x) = x - [x]$ is increasing in $(0, 1)$.

Prove that
$\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{\pi}{2}$

Construct the composition table for $+_5$ on set $S = \{0, 1, 2, 3, 4\}.$

Find the vector from the origin O to the centroid of the triangle whose vertices are (1, -1, 2), (2, 1, 3) and (-1, 2, -1).

The cartesian equation of a line AB are $\frac{2\text{x}-1}{\sqrt{3}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}.$ Find the direction cosines of a line parallel to AB.