What is the integral of x/((5-x^2)^7) ?

The integral of x/((5-x^2)^7) is 1/(12(5-x^2)^6)+C

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integral of x/((5-x^2)^7)

\int \frac{x}{( 5 - x ^{2} ) ^{7}} d x

\frac{1}{12 ( 5 - x ^{2} ) ^{6}} + C

Solution steps

\int \frac{x}{( 5 - x ^{2} ) ^{7}} d x

Apply u-substitution

: \int - \frac{1}{2 u ^{7}} d u

= \int - \frac{1}{2 u ^{7}} 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 ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= - \frac{1}{2} (- \frac{1}{6 u ^{6}})

Substitute back

u = 5 - x^{2}

= - \frac{1}{2} (- \frac{1}{6 ( 5 - x ^{2} ) ^{6}})

Simplify

- \frac{1}{2} (- \frac{1}{6 ( 5 - x ^{2} ) ^{6}}) : \frac{1}{12 ( 5 - x ^{2} ) ^{6}}

= \frac{1}{12 ( 5 - x ^{2} ) ^{6}}

Add a constant to the solution

= \frac{1}{12 ( 5 - x ^{2} ) ^{6}} + C

f (t) = sec^{2} (2 t)

y^{^{''}} + 3 y^{^{'}} - 4 y = 10 e^{x} - 250 cos (3 x)

n = 1 \sum \infty \frac{4 ^{n}}{n ^{4}}

\frac{d x}{d t} = - 3 x (\frac{1}{10} e^{- 2 t}), x (0) = \frac{4}{5}

\frac{\partial}{\partial y} (- 3 x^{2} + 7 x y - 4 y^{2} + x + y)

What is the integral of x/((5-x^2)^7) ?
The integral of x/((5-x^2)^7) is 1/(12(5-x^2)^6)+C