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

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

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

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

4 ln 1 + x^{2} + C

Solution steps

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

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

Simplify

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

= 4 ln 1 + x^{2}

Add a constant to the solution

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

\frac{d y}{d x} + y^{5} x + 6 y = 0

\int \frac{1}{x ^{2} 4 - x ^{2}} d x

\frac{d}{d x} (3 x e^{x} - cos (x))

f (t) = 5 - t

(ln (1 + 2 x))^{^{'}}

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