Question

integrate the function [2x/1 + x²].dx

Answer 1

$\bf{I=\int{\frac{2x}{1+x^2}},dx}$

Let (1 + x²) = t

differentiate both sides,

2x.dx = dt , put it in above Integration I

so, I = $\bf{I=\int{\frac{2x}{1+x^2}},dx}$

= $\bf{\int{\frac{dt}{t}}}$

= $\bf{log_e|t|+C}$

put t = (1 + x²)

I = $\bf{log_e|(1+x^2)|+C}$

Answer 2

Answer:

$I = \log|1+x^2| + C$

Step-by-step explanation:

Given,

$\displaystyle I = \int \dfrac{2x}{1+x^2}$

Derivative of (1+x²) is 2x.

since,

$\displaystyle \int\dfrac{\frac{dy}{dx}\left(f(x))}{f(x)} = \log|f(x)| + C$

Our integral will result into:

$\boxed{I = \log|1+x^2| + C}$