જો $\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = \frac{A \sqrt{x}}{\sqrt{1-x}} + \frac{B}{\sqrt{1-x}} + C$, જ્યાં $C$ એ વાસ્તવિક અચળાંક છે, તો $A+B$ ની કિંમત શોધો.

  • A
    $3$
  • B
    $0$
  • C
    $1$
  • D
    $2$

Explore More

Similar Questions

$\int {\left( {\sin \left( {101x} \right).{{\sin }^{99}}x} \right)} dx = \frac{{\sin \left( {100x} \right){{\left( {\sin x} \right)}^\lambda }}}{\mu } + C$ જ્યાં $C$ એ સંકલનનો અચળાંક છે,તો $\frac{\lambda }{\mu }$ ની કિંમત શોધો.

જો $\int {\frac{{\cos 4x + 1}}{{\cot x - \tan x}}} dx = k\,\cos 4x + c$ હોય,તો:

Difficult
View Solution

ધારો કે $I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}$. જો $I(37) - I(24) = \frac{1}{4} \left( \frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}} \right)$,જ્યાં $b, c \in \mathbb{N}$,તો $3(b+c)$ ની કિંમત શોધો.

$\int {(1 + x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}\,dx} = $

Difficult
View Solution

જો $\int \frac{1}{\cos 4x \cos 2x} dx = \frac{1}{2\sqrt{2}} \log \left(\frac{1+f(x)}{1-f(x)}\right) - \frac{1}{2} \log g(x) + C$ હોય,તો $g\left(\frac{\pi}{6}\right) - \sqrt{2} f\left(\frac{\pi}{6}\right)$ ની કિંમત શોધો.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo