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

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

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

integral

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

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

Solution steps

\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

sin^{3} (2 x + 5)

3 x^{2} (x^{3} + 1)^{10}

x^{2} lo g_{10} (1 + x)

x^{2} e^{l n (x) + x}

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

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