What is the integral of 8/(xsqrt(4x^2-1)) ?

The integral of 8/(xsqrt(4x^2-1)) is 8arctan(sqrt(4x^2-1))+C

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integral of 8/(xsqrt(4x^2-1))

\int \frac{8}{x 4 x ^{2} - 1} d x

8 arctan (4 x^{2} - 1) + C

Solution steps

\int \frac{8}{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

= 8 \cdot \int \frac{1}{x 4 x ^{2} - 1} d x

Apply u-substitution

= 8 \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 8 arctan (u)

Substitute back

u = 4 x^{2} - 1

= 8 arctan (4 x^{2} - 1)

Add a constant to the solution

= 8 arctan (4 x^{2} - 1) + C

What is the integral of 8/(xsqrt(4x^2-1)) ?
The integral of 8/(xsqrt(4x^2-1)) is 8arctan(sqrt(4x^2-1))+C