What is the integral of ((1+x)^2)/(1+x^2) ?

The integral of ((1+x)^2)/(1+x^2) is ln|1+x^2|+x+C

Popular >

integral of ((1+x)^2)/(1+x^2)

\int \frac{( 1 + x ) ^{2}}{1 + x ^{2}} d x

ln 1 + x^{2} + x + C

Solution steps

\int \frac{( 1 + x ) ^{2}}{1 + x ^{2}} d x

\frac{( 1 + x ) ^{2}}{1 + x ^{2}} = \frac{2 x}{1 + x ^{2}} + 1

= \int \frac{2 x}{1 + x ^{2}} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x}{1 + x ^{2}} d x + \int 1 d x

\int \frac{2 x}{1 + x ^{2}} d x = ln 1 + x^{2}

\int 1 d x = x

= ln 1 + x^{2} + x

Add a constant to the solution

= ln 1 + x^{2} + x + C

\frac{d y}{d t} + 3 y = 13 sin (2 t)

\int \frac{2 x + 1}{x ( x + 2 ) ^{3}} d x

2 x y d x + (x^{2} - y) d y = 0

\frac{\partial}{\partial x} (\frac{64}{x ^{2}})

\frac{\partial}{\partial x} (ln (1 + x^{2}))

What is the integral of ((1+x)^2)/(1+x^2) ?
The integral of ((1+x)^2)/(1+x^2) is ln|1+x^2|+x+C