What is the integral of x(2x^2+1)^{-1} ?

The integral of x(2x^2+1)^{-1} is 1/4 ln|2x^2+1|+C

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integral of x(2x^2+1)^{-1}

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

\frac{1}{4} ln 2 x^{2} + 1 + C

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} + 1

= \frac{1}{4} ln 2 x^{2} + 1

Add a constant to the solution

= \frac{1}{4} ln 2 x^{2} + 1 + C

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What is the integral of x(2x^2+1)^{-1} ?
The integral of x(2x^2+1)^{-1} is 1/4 ln|2x^2+1|+C