$\int \frac{d x}{2 \sqrt{x} (1 + x)} =$

\frac{1}{2} tan (\sqrt{x}) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan^{- 1} (\sqrt{x}) + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 tan^{- 1} (\sqrt{x}) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $tan^{- 1} (\sqrt{x}) + C$
$\begin{matrix} \int \frac{d x}{2 \sqrt{x} (1 + x)} P u t, \sqrt{x} = z \frac{1}{2 \sqrt{x}} d x = d z \frac{d x}{\sqrt{x}} = 2 d z s o, \int \frac{d x}{2 \sqrt{x} (1 + x)} = \int \frac{2 d z}{2 (1 + z^{2})} = \int \frac{d z}{(1 + z^{2})} = {tan}^{- 1} (z) + c = {tan}^{- 1} (\sqrt{x}) + c \end{matrix}$

Suggest Corrections

Similar questions

y = log ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ {(\frac{1 + x}{1 - x})}^{\frac{1}{4}} ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ - \frac{1}{2} {tan}^{- 1} (x)

\frac{d y}{d x} =

$\int \frac{x^{2} + 1}{x^{4} + 1} d x$ will be equal to which of the following

If \int cot (2 {tan}^{- 1} \sqrt{\frac{\sqrt{1 + \sqrt{x}} - x^{\frac{1}{4}}}{\sqrt{1 + \sqrt{x}} + x^{\frac{1}{4}}}}) d x = \frac{q . x^{\frac{p}{4}}}{p} + C, x > 0

p

q

C

I_{1} = \int \frac{d x}{e^{x} + 8 e^{- x} + 4 e^{- 3 x}}

I_{2} = \int \frac{d x}{e^{3 x} + 8 e^{x} + 4 e^{- x}}

I = I_{1} - 2 I_{2}

If \int cot (2 {tan}^{- 1} \sqrt{\frac{\sqrt{1 + \sqrt{x}} - x^{\frac{1}{4}}}{\sqrt{1 + \sqrt{x}} + x^{\frac{1}{4}}}}) d x = \frac{q . x^{\frac{p}{4}}}{p} + C, x > 0

p

q

C

Join BYJU'S Learning Program