What is the integral of (8x)/(2x^2+1) ?

The integral of (8x)/(2x^2+1) is 2ln|2x^2+1|+C

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integral of (8x)/(2x^2+1)

\int \frac{8 x}{2 x ^{2} + 1} d x

2 ln 2 x^{2} + 1 + C

Solution steps

\int \frac{8 x}{2 x ^{2} + 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{2 x ^{2} + 1} d x

Apply u-substitution

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

= 8 \cdot \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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} + 1

= 8 \cdot \frac{1}{4} ln 2 x^{2} + 1

Simplify

8 \cdot \frac{1}{4} ln 2 x^{2} + 1 : 2 ln 2 x^{2} + 1

= 2 ln 2 x^{2} + 1

Add a constant to the solution

= 2 ln 2 x^{2} + 1 + C

y = (x + 1)^{3}

\int \frac{x}{x ^{2} - 2} d x

\int \frac{1}{16 + 25 x ^{2}} d x

\int \frac{x + 1}{x} d x

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

What is the integral of (8x)/(2x^2+1) ?
The integral of (8x)/(2x^2+1) is 2ln|2x^2+1|+C