What is the integral of (16x)/(8x^2+2) ?

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

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integral of (16x)/(8x^2+2)

\int \frac{16 x}{8 x ^{2} + 2} d x

ln 8 x^{2} + 2 + C

Solution steps

\int \frac{16 x}{8 x ^{2} + 2} d x

Take the constant out:

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

= 16 \cdot \int \frac{x}{8 x ^{2} + 2} d x

Apply u-substitution

: \int \frac{1}{16 u} d u

= 16 \cdot \int \frac{1}{16 u} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 \cdot \frac{1}{16} ln ∣ u ∣

Substitute back

u = 8 x^{2} + 2

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

Simplify

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

= ln 8 x^{2} + 2

Add a constant to the solution

= ln 8 x^{2} + 2 + C

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