Question

Integral x(root of x²+1)dx

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\longrightarrow \: \tt \: Let \: I \: = \int \: x \sqrt{ {x}^{2} + 1 } dx$

$\longrightarrow \: \tt \: \red{Put \: \sqrt{ {x}^{2} + 1} = y}$

☆ On differentiate both sides w. r. t. x, we get

$\longrightarrow \: \tt \: \dfrac{d}{dx} \sqrt{ {x}^{2} + 1} = \dfrac{dy}{dx}$

$\longrightarrow \: \tt \: \dfrac{1}{2 \sqrt{ {x}^{2} + 1 } } \times 2x = \dfrac{dy}{dx}$

$\tt\implies \:\dfrac{x}{y} = \dfrac{dy}{dx}$

$\tt\implies \:x \: dx = y \: dy$

$\longrightarrow \: \tt \: So, \: I \: = \int \: y \times ydy$

$\longrightarrow \: \tt \: I = \int \: {y}^{2} dy$

$\longrightarrow \: \tt \: I = \dfrac{ {y}^{3} }{3} + c$

$\longrightarrow \: \tt \: I = \dfrac{ {( {x}^{2} + 1) }^{ \frac{3}{2} } }{3} + c$