What is the integral of (4x)/(x^4-12x^2+45) ?

The integral of (4x)/(x^4-12x^2+45) is 2/(sqrt(33))arctan((x^2)/(sqrt(33)))+C

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integral of (4x)/(x^4-12x^2+45)

\int \frac{4 x}{x ^{4} - 12 x ^{2} + 45} d x

\frac{2}{33} arctan (\frac{x ^{2}}{33}) + C

Solution steps

\int \frac{4 x}{x ^{4} - 12 x ^{2} + 45} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{4} - 12 x ^{2} + 45} d x

Apply u-substitution

: \int \frac{1}{4 u ( u + 45 - 12 u )} d u

= 4 \cdot \int \frac{1}{4 u ( u + 45 - 12 u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 45 - 12 u )} d u

Apply exponent rule:

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

\frac{1}{u ( u + 45 - 12 u )} = u^{- \frac{1}{2}} \frac{1}{( u + 45 - 12 )}

= 4 \cdot \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 45 - 12} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{u + 45 - 12} : \frac{1}{u ( u + 33 )}

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 33 )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 33} d v

= 4 \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 33} d v

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 33} d v

Apply Integral Substitution

= 4 \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{33 ( w ^{2} + 1 )} d w

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{33} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= 4 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{33} arctan (w)

Substitute back

= 4 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{33} arctan (\frac{x ^{4}}{33})

Simplify

4 \cdot \frac{1}{4} \cdot 2 \cdot \frac{1}{33} arctan (\frac{x ^{4}}{33}) : \frac{2}{33} arctan (\frac{x ^{2}}{33})

= \frac{2}{33} arctan (\frac{x ^{2}}{33})

Add a constant to the solution

= \frac{2}{33} arctan (\frac{x ^{2}}{33}) + C

f (x) = 4 x^{2}, at x = 1

\frac{\partial}{\partial x} (x^{9 y})

\frac{\partial}{\partial u} (u \cdot cos (v))

f (x) = arccos (x)

y = 2 sin (π x - y)

What is the integral of (4x)/(x^4-12x^2+45) ?
The integral of (4x)/(x^4-12x^2+45) is 2/(sqrt(33))arctan((x^2)/(sqrt(33)))+C