What is the integral of (7x^2)/((36+x^2)^2) ?

The integral of (7x^2)/((36+x^2)^2) is (7(x^2arctan(x/6)-6x+36arctan(x/6)))/(12(36+x^2))+C

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integral of (7x^2)/((36+x^2)^2)

\int \frac{7 x ^{2}}{( 36 + x ^{2} ) ^{2}} d x

\frac{7 ( x ^{2} arctan ( \frac{x}{6} ) - 6 x + 36 arctan ( \frac{x}{6} ) )}{12 ( 36 + x ^{2} )} + C

Solution steps

\int \frac{7 x ^{2}}{( 36 + x ^{2} ) ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{x ^{2}}{( 36 + x ^{2} ) ^{2}} d x

Apply Integral Substitution

= 7 \cdot \int \frac{u ^{2}}{6 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 7 \cdot \frac{1}{6} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

Take the partial fraction of

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

= 7 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 \cdot \frac{1}{6} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

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

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

= 7 \cdot \frac{1}{6} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{x}{6}

= 7 \cdot \frac{1}{6} (arctan (\frac{x}{6}) - \frac{1}{2} (arctan (\frac{x}{6}) + \frac{\frac{x}{6}}{1 + ( \frac{x}{6} ) ^{2}}))

Simplify

7 \cdot \frac{1}{6} (arctan (\frac{x}{6}) - \frac{1}{2} (arctan (\frac{x}{6}) + \frac{\frac{x}{6}}{1 + ( \frac{x}{6} ) ^{2}})) : \frac{7 ( x ^{2} arctan ( \frac{x}{6} ) - 6 x + 36 arctan ( \frac{x}{6} ) )}{12 ( 36 + x ^{2} )}

= \frac{7 ( x ^{2} arctan ( \frac{x}{6} ) - 6 x + 36 arctan ( \frac{x}{6} ) )}{12 ( 36 + x ^{2} )}

Add a constant to the solution

= \frac{7 ( x ^{2} arctan ( \frac{x}{6} ) - 6 x + 36 arctan ( \frac{x}{6} ) )}{12 ( 36 + x ^{2} )} + C

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\frac{\partial}{\partial z} (e^{x y} + z)

\frac{d}{d x} ((x + 2)^{2} + (10 - 3 x)^{2})

n = 1 \sum \infty 1 + \frac{1}{n + 1}

x \to 0 lim (\frac{( 2 - x ) ^{3} - 8}{x})

What is the integral of (7x^2)/((36+x^2)^2) ?
The integral of (7x^2)/((36+x^2)^2) is (7(x^2arctan(x/6)-6x+36arctan(x/6)))/(12(36+x^2))+C