What is the integral of (18)/((81x^2+1)^2) ?

The integral of (18)/((81x^2+1)^2) is arctan(9x)+1/2 sin(2arctan(9x))+C

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integral of (18)/((81x^2+1)^2)

\int \frac{18}{( 81 x ^{2} + 1 ) ^{2}} d x

arctan (9 x) + \frac{1}{2} sin (2 arctan (9 x)) + C

Solution steps

\int \frac{18}{( 81 x ^{2} + 1 ) ^{2}} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{( 81 x ^{2} + 1 ) ^{2}} d x

Apply Integral Substitution

= 18 \cdot \int \frac{1}{9 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= 18 \cdot \frac{1}{9} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

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

= 18 \cdot \frac{1}{9} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 18 \cdot \frac{1}{9} \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

= 18 \cdot \frac{1}{9} \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

= 18 \cdot \frac{1}{9} \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)

= 18 \cdot \frac{1}{9} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 18 \cdot \frac{1}{9} \cdot \frac{1}{2} (arctan (9 x) + \frac{1}{2} sin (2 arctan (9 x)))

Simplify

18 \cdot \frac{1}{9} \cdot \frac{1}{2} (arctan (9 x) + \frac{1}{2} sin (2 arctan (9 x))) : arctan (9 x) + \frac{1}{2} sin (2 arctan (9 x))

= arctan (9 x) + \frac{1}{2} sin (2 arctan (9 x))

Add a constant to the solution

= arctan (9 x) + \frac{1}{2} sin (2 arctan (9 x)) + C

h (t) = t (1 - t^{2})

x \cdot e^{(- 4 x)}

(- 7.2) (9.6)

\int_{1}^{2} \frac{4}{x ^{3} + 4 x} d x

\frac{d}{d x} (csc (8 x))

What is the integral of (18)/((81x^2+1)^2) ?
The integral of (18)/((81x^2+1)^2) is arctan(9x)+1/2 sin(2arctan(9x))+C