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

The integral of (8x)/(x^2+1) is 4ln|x^2+1|+C

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

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

4 ln x^{2} + 1 + C

Solution steps

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

Apply u-substitution

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

= 8 \cdot \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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

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

Simplify

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

= 4 ln x^{2} + 1

Add a constant to the solution

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

\frac{d ^{3}}{d x ^{3}} (cos (2 x))

f (x) = \frac{x ^{2} e ^{x}}{x ^{2} + e ^{x}}

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

y^{^{'}} = \frac{x ^{6}}{y}

\int x^{2} + 4 d x

What is the integral of (8x)/(x^2+1) ?
The integral of (8x)/(x^2+1) is 4ln|x^2+1|+C