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

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

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integral of p/(p^2+1)

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

\frac{1}{2} ln p^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = p^{2} + 1

= \frac{1}{2} ln p^{2} + 1

Add a constant to the solution

= \frac{1}{2} ln p^{2} + 1 + C

\int \frac{arcsin ( x )}{x ^{2}} d x

\frac{\partial}{\partial x} (x + 2)

9 x, x^{3}

\int u^{- 5} (8 - u^{5} + u^{10}) d u

n \to \infty lim ((n + 1) x)

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