Question

integral of (x+4/1+x^2)dx​

Answer 1

Answer:

$\sf I=\dfrac{1}{2}\times log|1+x^2|+ 4\:tan^{-1}x+C$

Step-by-step explanation:

Given:

$\sf \dfrac{x+4}{1+x^2}$

To Find:

$\displaystyle \sf \int\limits {\dfrac{x+4}{1+x^2} } \, dx$

Solution:

The given integral can be split into,

$\displaystyle \sf \int\limits {\dfrac{x}{x^2+1} } \, dx +\int\limits {\dfrac{4}{x^2+1} } \, dx$

$\implies \displaystyle \sf \int\limits {\dfrac{x}{x^2+1} } \, dx +4\int\limits {\dfrac{1}{x^2+1} } \, dx$

Now let x² + 1 = t

Differentiate on both sides with respect to x,

2x dx = dt

x dx = dt/2

Substituting this above we get,

$\displaystyle \sf \int\limits {\dfrac{1}{2t} } \, dt+4\times tan^{-1} x+C$

$\implies \displaystyle \sf \dfrac{1}{2} \int\limits {\dfrac{1}{t} } \, dt+4\times tan^{-1} x+C$

$\implies \sf \dfrac{1}{2}\times log|t|+ 4\:tan^{-1}x+C$

Give back the value of t,

$\sf \dfrac{1}{2}\times log|1+x^2|+ 4\:tan^{-1}x+C$

This is the required integral.