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

The integral of (9/(x^2+9))^2 is 3/2 (arctan(x/3)+1/2 sin(2arctan(x/3)))+C

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

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

\frac{3}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3}))) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{3}{( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= 3 \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 3 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 3 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 3 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 3 \cdot \frac{1}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

Simplify

3 \cdot \frac{1}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3}))) : \frac{3}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

= \frac{3}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3})))

Add a constant to the solution

= \frac{3}{2} (arctan (\frac{x}{3}) + \frac{1}{2} sin (2 arctan (\frac{x}{3}))) + C

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\int 3 x 5 - 2 x^{2} d x

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What is the integral of (9/(x^2+9))^2 ?
The integral of (9/(x^2+9))^2 is 3/2 (arctan(x/3)+1/2 sin(2arctan(x/3)))+C