What is the integral of-(32x)/(4x^2+1) ?

The integral of-(32x)/(4x^2+1) is -4ln|4x^2+1|+C

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integral of-(32x)/(4x^2+1)

\int - \frac{32 x}{4 x ^{2} + 1} d x

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

Solution steps

\int - \frac{32 x}{4 x ^{2} + 1} d x

Take the constant out:

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

= - 32 \cdot \int \frac{x}{4 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= - 32 \cdot \int \frac{1}{8 u} d u

Take the constant out:

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

= - 32 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 32 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 x^{2} + 1

= - 32 \cdot \frac{1}{8} ln 4 x^{2} + 1

Simplify

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

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

Add a constant to the solution

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

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What is the integral of-(32x)/(4x^2+1) ?
The integral of-(32x)/(4x^2+1) is -4ln|4x^2+1|+C