What is the integral of x/(x^2sqrt(9x^4-25)) ?

The integral of x/(x^2sqrt(9x^4-25)) is 1/10 arctan(1/5 sqrt(9x^4-25))+C

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integral of x/(x^2sqrt(9x^4-25))

\int \frac{x}{x ^{2} 9 x ^{4} - 25} d x

\frac{1}{10} arctan (\frac{1}{5} 9 x^{4} - 25) + C

Solution steps

\int \frac{x}{x ^{2} 9 x ^{4} - 25} d x

Cancel the common factor:

x

= \int \frac{1}{x 9 x ^{4} - 25} d x

Apply u-substitution

= \int \frac{1}{4 u 9 u - 25} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u 9 u - 25} d u

Apply u-substitution

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 25} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 25} d v

Apply Integral Substitution

= \frac{1}{4} \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}{4} \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}{4} \cdot 2 \cdot \frac{1}{5} arctan (w)

Substitute back

= \frac{1}{4} \cdot 2 \cdot \frac{1}{5} arctan (\frac{9 x ^{4} - 25}{5})

Simplify

\frac{1}{4} \cdot 2 \cdot \frac{1}{5} arctan (\frac{9 x ^{4} - 25}{5}) : \frac{1}{10} arctan (\frac{1}{5} 9 x^{4} - 25)

= \frac{1}{10} arctan (\frac{1}{5} 9 x^{4} - 25)

Add a constant to the solution

= \frac{1}{10} arctan (\frac{1}{5} 9 x^{4} - 25) + C

What is the integral of x/(x^2sqrt(9x^4-25)) ?
The integral of x/(x^2sqrt(9x^4-25)) is 1/10 arctan(1/5 sqrt(9x^4-25))+C