$\int_{}^{}\frac{x^{49} an^{- 1}{(x^{50})}}{(1 + x^{100})}dx = k\left\lbrack an^{- 1}\left( x^{50} ight) ight brack^{2} + c$, then $k$ is equal to

Question

$\int_{}^{}\frac{x^{49}	an^{- 1}{(x^{50})}}{(1 + x^{100})}dx = k\left\lbrack 	an^{- 1}\left( x^{50} ight) ightbrack^{2} + c$, then $k$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Let $I = \int_{}^{}x^{49}\frac{	an^{- 1}{(x^{50})}}{\left( 1 + x^{50} ight)^{2}}dx$
Put $x^{50} = t\ \  \Rightarrow \ 50x^{49}\ dx = dt$
$	herefore\ I = \frac{1}{50}\ \ \int_{}^{}\frac{	an^{- 1}t}{1 + t^{2}}dt$
Again, put $	an^{- 1}t = u\  \Rightarrow \ \frac{1}{1 + t^{2}}dt = du$
$	herefore\ \ I = \frac{1}{50}\int_{}^{}{u\ du = \frac{u^{2}}{100}} + c$
$= \frac{\left( 	an^{- 1}x^{50} ight)^{2}}{100} + c$
But $I = k\left( 	an^{- 1}x^{50} ight)^{2} + c$ [given]
$	herefore\ k = \frac{1}{100}$

Q. $\int_{}^{}\frac{x^{49}\tan^{- 1}{(x^{50})}}{(1 + x^{100})}dx = k\left\lbrack \tan^{- 1}\left( x^{50} \right) \right\rbrack^{2} + c$, then $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?