What is the integral of (x^3)/((2+x^4)^9) ?

The integral of (x^3)/((2+x^4)^9) is -1/(32(2+x^4)^8)+C

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integral of (x^3)/((2+x^4)^9)

\int \frac{x ^{3}}{( 2 + x ^{4} ) ^{9}} d x

- \frac{1}{32 ( 2 + x ^{4} ) ^{8}} + C

Solution steps

\int \frac{x ^{3}}{( 2 + x ^{4} ) ^{9}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{9}} d u

= \int \frac{1}{4 u ^{9}} 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}} d u

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= \frac{1}{4} (- \frac{1}{8 u ^{8}})

Substitute back

u = 2 + x^{4}

= \frac{1}{4} (- \frac{1}{8 ( 2 + x ^{4} ) ^{8}})

Simplify

\frac{1}{4} (- \frac{1}{8 ( 2 + x ^{4} ) ^{8}}) : - \frac{1}{32 ( 2 + x ^{4} ) ^{8}}

= - \frac{1}{32 ( 2 + x ^{4} ) ^{8}}

Add a constant to the solution

= - \frac{1}{32 ( 2 + x ^{4} ) ^{8}} + C

y^{^{'}} = 2 y - 3 e^{- t}

x \to 0 lim (sin (\frac{2}{x}))

f (x) = 2 x + x

\frac{d y}{d x} + 3 y = 18

\int \frac{1}{( x ^{2} + 6 x + 6 ) ^{2}} d x

What is the integral of (x^3)/((2+x^4)^9) ?
The integral of (x^3)/((2+x^4)^9) is -1/(32(2+x^4)^8)+C