What is the integral of x/(x^2sqrt(x^2-4)) ?

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

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integral of x/(x^2sqrt(x^2-4))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u u - 4} d u

= \int \frac{1}{2 u u - 4} 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 u - 4} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 4} d v

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 4} 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} + 4} d v

Apply Integral Substitution

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

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} arctan (\frac{x ^{2} - 4}{2})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{2} arctan (\frac{x ^{2} - 4}{2}) : \frac{1}{2} arctan (\frac{1}{2} x^{2} - 4)

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

Add a constant to the solution

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

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