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

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

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

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

3 ln 1 + x^{2} + C

Solution steps

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

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

Simplify

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

= 3 ln 1 + x^{2}

Add a constant to the solution

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

x^{2} + x y - y^{2} = - 4, (2, 4)

(2 x - 1)^{3}

\int 3 + x d x

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

\int \frac{4 x ^{2}}{1 + x ^{3}} d x

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