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

Popular >

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

: \int \frac{1}{u ^{2} + 1} d u

= 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

\int x^{4} (x^{5} - 3)^{3} d x

\int 7 x^{3} cos (2 + x^{4}) - 8 x^{3} e^{2 + x^{4}} d x

\int (e^{\frac{x}{a}} + e^{- \frac{x}{a}})^{2} d x

\int 5^{\frac{x}{2}} d x

\int (3 t^{2} i + 4 t j - 8 t^{3} k) d t

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