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

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

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

integral

\frac{v + 1}{v ^{2} + 1}

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

Solution steps

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

Expand

\frac{v + 1}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{1}{v ^{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{v}{v ^{2} + 1} d v + \int \frac{1}{v ^{2} + 1} d v

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

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

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

Add a constant to the solution

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

7 x^{6}

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

sin (x^{2} + 1)

2 + 5 x

x (4 - x)^{\frac{1}{2}}

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