Question

Question from Integral Calculas

Class 12th for Maths Genius and Moderators.

Solve it

Don't Spam
$let \: f(x) = \large{ \int} \frac{ \sqrt{x} }{ {(1 + x)}^{2} } dx(x \geqslant 0)$
Then f(3) - f(1) is equals to ?​

Answer 1

ANSWER:-)

$\dfrac{ \pi}{12} + \dfrac{1}{2} - \dfrac{ \sqrt{3} }{4}$

EXPLANATION:-)

Given that

$f(x) = \int \frac{ \sqrt{x} }{ {(1 + x)}^{2} }dx \\ \\ put \: x = {tan}^{2} y \\ \implies dx = 2tan \: y \: {sec}^{2} \: ydy \\ \\ \therefore f(x) = \int \frac{2tan {}^{2} y. {sec}^{2} y}{ {sec}^{4} y} dy \\ \\ = \int2 {sin}^{2} ydy \\ \\ = \int(1 - cos2y)dy \\ \\ = y-\frac{sin2y}{2} + C$

$= y - \dfrac{tan \: y}{1 + {tan}^{2} y} + C \\ \\ = {tan}^{ - 1} \sqrt{x} - \frac{ \sqrt{x} }{1 + x} + C$

NOW:-)

$f(3) - f(1) \\ \\ = ( {tan}^{ - 1} \sqrt{3} - \frac{ \sqrt{3} }{1 + 3} + C) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - ( {tan}^{ - 1} 1 - \frac{1}{1 + 1} + C) \\ \\ = \frac{ \pi}{3} - \frac{ \sqrt{3} }{4} - \frac{ \pi}{4} + \frac{1}{2} \\ \\ = \frac{ \pi}{12} + \frac{1}{2} - \frac{ \sqrt{3} }{4}$

Which is the required Answer of this Question

Answer 2

Answer:

• Given that,

$\huge \bf \frac{f′(x)}{2}= \frac{{e }^{f(x)}}{x}</p><p></p><p></p><p>$

$\huge{ \red{ \tt⟹f′(x).{e }^{ f(x)}=2x}}$

$\large {\pink{⟹∫f′(x).{e }^{f(x)}dx=∫2xdx}</p><p>}$

$\blue {\huge{⟹{e }^{f(x)}={x }^{2} +c}}$

$\huge{ \red{∵x=1,f(x)=f(1)=1}}$

$\tt{⟹{e }^{ f(x)}={x }^{2} +e−1}$

$\huge {\red{ \cal{⟹f(x)=log _ e({x }^{2} +e−1)}}}$

When, f(x)=f(x) then the number of solutions are infinite.