What is the integral of (10X)/(X^4-13X^2+36) ?

The integral of (10X)/(X^4-13X^2+36) is 5/(sqrt(23))arctan((X^2)/(sqrt(23)))+C

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integral of (10X)/(X^4-13X^2+36)

\int \frac{10 X}{X ^{4} - 13 X ^{2} + 36} d X

\frac{5}{23} arctan (\frac{X ^{2}}{23}) + C

Solution steps

\int \frac{10 X}{X ^{4} - 13 X ^{2} + 36} d X

Take the constant out:

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

= 10 \cdot \int \frac{X}{X ^{4} - 13 X ^{2} + 36} d X

Apply u-substitution

: \int \frac{1}{4 u ( u + 36 - 13 u )} d u

= 10 \cdot \int \frac{1}{4 u ( u + 36 - 13 u )} d u

Take the constant out:

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

= 10 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 36 - 13 u )} d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u ( u + 36 - 13 u )} = u^{- \frac{1}{2}} \frac{1}{( u + 36 - 13 )}

= 10 \cdot \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 36 - 13} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{u + 36 - 13} : \frac{1}{u ( u + 23 )}

= 10 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 23 )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 23} d v

= 10 \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 23} d v

Take the constant out:

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

= 10 \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 23} d v

Apply Integral Substitution

= 10 \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{23 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 10 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{23} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 10 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{23} arctan (w)

Substitute back

= 10 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{23} arctan (\frac{X ^{4}}{23})

Simplify

10 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{23} arctan (\frac{X ^{4}}{23}) : \frac{5}{23} arctan (\frac{X ^{2}}{23})

= \frac{5}{23} arctan (\frac{X ^{2}}{23})

Add a constant to the solution

= \frac{5}{23} arctan (\frac{X ^{2}}{23}) + C

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What is the integral of (10X)/(X^4-13X^2+36) ?
The integral of (10X)/(X^4-13X^2+36) is 5/(sqrt(23))arctan((X^2)/(sqrt(23)))+C