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))

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

\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

y = ln (x + 6)

\int (sin (x))^{4} d x

x^{(\frac{1}{3})}, 8

\int 2 x x^{2} + 4 d x

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