What is the integral of (6x)/(x^2+1) ?

The integral of (6x)/(x^2+1) is 3ln|x^2+1|+C

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integral of (6x)/(x^2+1)

\int \frac{6 x}{x ^{2} + 1} d x

3 ln x^{2} + 1 + C

Solution steps

\int \frac{6 x}{x ^{2} + 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{x ^{2} + 1} d x

Apply u-substitution

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

= 6 \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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

= 6 \cdot \frac{1}{2} ln x^{2} + 1

Simplify

6 \cdot \frac{1}{2} ln x^{2} + 1 : 3 ln x^{2} + 1

= 3 ln x^{2} + 1

Add a constant to the solution

= 3 ln x^{2} + 1 + C

\frac{d ^{2}}{d x ^{2}} ((6 x^{2} + 8)^{6})

n = 1 \sum \infty \frac{4}{π ^{n}}

\int x^{2} 6 x + 3 d x

f (x) = \frac{1}{5 x + 7}

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

What is the integral of (6x)/(x^2+1) ?
The integral of (6x)/(x^2+1) is 3ln|x^2+1|+C