What is the integral of (4x)/(16x^2+1) ?

The integral of (4x)/(16x^2+1) is 1/8 ln|16x^2+1|+C

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integral of (4x)/(16x^2+1)

\int \frac{4 x}{16 x ^{2} + 1} d x

\frac{1}{8} ln 16 x^{2} + 1 + C

Solution steps

\int \frac{4 x}{16 x ^{2} + 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{16 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{32 u} d u

= 4 \cdot \int \frac{1}{32 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{32} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{32} ln ∣ u ∣

Substitute back

u = 16 x^{2} + 1

= 4 \cdot \frac{1}{32} ln 16 x^{2} + 1

Simplify

4 \cdot \frac{1}{32} ln 16 x^{2} + 1 : \frac{1}{8} ln 16 x^{2} + 1

= \frac{1}{8} ln 16 x^{2} + 1

Add a constant to the solution

= \frac{1}{8} ln 16 x^{2} + 1 + C

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What is the integral of (4x)/(16x^2+1) ?
The integral of (4x)/(16x^2+1) is 1/8 ln|16x^2+1|+C