What is the integral of 1/(x(64x^2-1)^{1/2)} ?

The integral of 1/(x(64x^2-1)^{1/2)} is arctan((64x^2-1)^{1/2})+C

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integral of 1/(x(64x^2-1)^{1/2)}

\int \frac{1}{x ( 64 x ^{2} - 1 ) ^{\frac{1}{2}}} d x

arctan ((64 x^{2} - 1)^{\frac{1}{2}}) + C

Solution steps

\int \frac{1}{x ( 64 x ^{2} - 1 ) ^{\frac{1}{2}}} d x

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = (64 x^{2} - 1)^{\frac{1}{2}}

= arctan ((64 x^{2} - 1)^{\frac{1}{2}})

Add a constant to the solution

= arctan ((64 x^{2} - 1)^{\frac{1}{2}}) + C

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What is the integral of 1/(x(64x^2-1)^{1/2)} ?
The integral of 1/(x(64x^2-1)^{1/2)} is arctan((64x^2-1)^{1/2})+C