What is the integral of (2x)/(16x^2-25) ?

The integral of (2x)/(16x^2-25) is 1/16 ln|16x^2-25|+C

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integral of (2x)/(16x^2-25)

\int \frac{2 x}{16 x ^{2} - 25} d x

\frac{1}{16} ln 16 x^{2} - 25 + C

Solution steps

\int \frac{2 x}{16 x ^{2} - 25} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{16 x ^{2} - 25} d x

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{32} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{32} ln ∣ u ∣

Substitute back

u = 16 x^{2} - 25

= 2 \cdot \frac{1}{32} ln 16 x^{2} - 25

Simplify

2 \cdot \frac{1}{32} ln 16 x^{2} - 25 : \frac{1}{16} ln 16 x^{2} - 25

= \frac{1}{16} ln 16 x^{2} - 25

Add a constant to the solution

= \frac{1}{16} ln 16 x^{2} - 25 + C

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What is the integral of (2x)/(16x^2-25) ?
The integral of (2x)/(16x^2-25) is 1/16 ln|16x^2-25|+C