What is the integral of x/(x^2sqrt(x^2-16)) ?

The integral of x/(x^2sqrt(x^2-16)) is 1/4 arctan(1/4 sqrt(x^2-16))+C

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integral of x/(x^2sqrt(x^2-16))

\int \frac{x}{x ^{2} x ^{2} - 16} d x

\frac{1}{4} arctan (\frac{1}{4} x^{2} - 16) + C

Solution steps

\int \frac{x}{x ^{2} x ^{2} - 16} d x

Apply u-substitution

: \int \frac{1}{2 u u - 16} d u

= \int \frac{1}{2 u u - 16} 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 - 16} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 16} d v

= \frac{1}{2} \cdot \int \frac{2}{v ^{2} + 16} 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} + 16} d v

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{4 ( 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}{4} \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}{4} arctan (w)

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{4} arctan (\frac{x ^{2} - 16}{4})

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{4} arctan (\frac{x ^{2} - 16}{4}) : \frac{1}{4} arctan (\frac{1}{4} x^{2} - 16)

= \frac{1}{4} arctan (\frac{1}{4} x^{2} - 16)

Add a constant to the solution

= \frac{1}{4} arctan (\frac{1}{4} x^{2} - 16) + C

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What is the integral of x/(x^2sqrt(x^2-16)) ?
The integral of x/(x^2sqrt(x^2-16)) is 1/4 arctan(1/4 sqrt(x^2-16))+C