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

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

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

integral

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

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

Solution steps

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

Expand

\frac{x + 1}{x ^{2} + 1} : \frac{x}{x ^{2} + 1} + \frac{1}{x ^{2} + 1}

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

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

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

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

Add a constant to the solution

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

\frac{1}{( 1 - y )}

(\frac{ln ( x )}{x})^{2}

\frac{x ^{3}}{( 1 - x ^{4} )}

sin^{2} (\frac{x}{2})

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

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