What is the integral of 1/(xsqrt(2x^2-5)) ?

The integral of 1/(xsqrt(2x^2-5)) is 1/(sqrt(5))arctan(sqrt(1/5 (2x^2-5)))+C

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integral of 1/(xsqrt(2x^2-5))

\int \frac{1}{x 2 x ^{2} - 5} d x

\frac{1}{5} arctan (\frac{1}{5} (2 x^{2} - 5)) + C

Solution steps

\int \frac{1}{x 2 x ^{2} - 5} d x

Apply u-substitution

: \int \frac{1}{2 u 2 u - 5} d u

= \int \frac{1}{2 u 2 u - 5} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u 2 u - 5} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 5} d v

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 5} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v ^{2} + 5} d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{5 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{5} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{5} arctan (w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{5} arctan (\frac{2 x ^{2} - 5}{5})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{5} arctan (\frac{2 x ^{2} - 5}{5}) : \frac{1}{5} arctan (\frac{1}{5} (2 x^{2} - 5))

= \frac{1}{5} arctan (\frac{1}{5} (2 x^{2} - 5))

Add a constant to the solution

= \frac{1}{5} arctan (\frac{1}{5} (2 x^{2} - 5)) + C

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What is the integral of 1/(xsqrt(2x^2-5)) ?
The integral of 1/(xsqrt(2x^2-5)) is 1/(sqrt(5))arctan(sqrt(1/5 (2x^2-5)))+C