What is the integral of 1/((9x^2+16)^2) ?

The integral of 1/((9x^2+16)^2) is 1/384 (arctan(3/4 x)+1/2 sin(2arctan(3/4 x)))+C

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integral of 1/((9x^2+16)^2)

\int \frac{1}{( 9 x ^{2} + 16 ) ^{2}} d x

\frac{1}{384} (arctan (\frac{3}{4} x) + \frac{1}{2} sin (2 arctan (\frac{3}{4} x))) + C

Solution steps

\int \frac{1}{( 9 x ^{2} + 16 ) ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{192 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{192} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

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

= \frac{1}{192} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{192} \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

= \frac{1}{192} \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

= \frac{1}{192} \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)

= \frac{1}{192} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{192} \cdot \frac{1}{2} (arctan (\frac{3}{4} x) + \frac{1}{2} sin (2 arctan (\frac{3}{4} x)))

Simplify

\frac{1}{192} \cdot \frac{1}{2} (arctan (\frac{3}{4} x) + \frac{1}{2} sin (2 arctan (\frac{3}{4} x))) : \frac{1}{384} (arctan (\frac{3}{4} x) + \frac{1}{2} sin (2 arctan (\frac{3}{4} x)))

= \frac{1}{384} (arctan (\frac{3}{4} x) + \frac{1}{2} sin (2 arctan (\frac{3}{4} x)))

Add a constant to the solution

= \frac{1}{384} (arctan (\frac{3}{4} x) + \frac{1}{2} sin (2 arctan (\frac{3}{4} x))) + C

\frac{d}{d x} (2 x (12 - x)^{3})

ln (1 + 8 x^{2})

cos (\frac{1 - e ^{9 x}}{1 + e ^{9 x}})

x y + y^{2} + x^{2} - x^{2} \frac{d y}{d x} = 0

x \to + 0 lim (7)

What is the integral of 1/((9x^2+16)^2) ?
The integral of 1/((9x^2+16)^2) is 1/384 (arctan(3/4 x)+1/2 sin(2arctan(3/4 x)))+C