What is the integral of (12)/((36x^2+1)^2) ?

The integral of (12)/((36x^2+1)^2) is arctan(6x)+1/2 sin(2arctan(6x))+C

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integral of (12)/((36x^2+1)^2)

\int \frac{12}{( 36 x ^{2} + 1 ) ^{2}} d x

arctan (6 x) + \frac{1}{2} sin (2 arctan (6 x)) + C

Solution steps

\int \frac{12}{( 36 x ^{2} + 1 ) ^{2}} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{( 36 x ^{2} + 1 ) ^{2}} d x

Apply Integral Substitution

= 12 \cdot \int \frac{1}{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

= 12 \cdot \frac{1}{6} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

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

= 12 \cdot \frac{1}{6} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 12 \cdot \frac{1}{6} \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

= 12 \cdot \frac{1}{6} \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

= 12 \cdot \frac{1}{6} \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)

= 12 \cdot \frac{1}{6} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 12 \cdot \frac{1}{6} \cdot \frac{1}{2} (arctan (6 x) + \frac{1}{2} sin (2 arctan (6 x)))

Simplify

12 \cdot \frac{1}{6} \cdot \frac{1}{2} (arctan (6 x) + \frac{1}{2} sin (2 arctan (6 x))) : arctan (6 x) + \frac{1}{2} sin (2 arctan (6 x))

= arctan (6 x) + \frac{1}{2} sin (2 arctan (6 x))

Add a constant to the solution

= arctan (6 x) + \frac{1}{2} sin (2 arctan (6 x)) + C

ln (3 x^{2})

h (x) = 3 e^{x} - 7 x, (0, 3)

\frac{d}{d x} (e^{t a n (3 x)})

(9, 2), (- 2, 6)

\frac{d}{d x} (\frac{2}{x} - 1)

What is the integral of (12)/((36x^2+1)^2) ?
The integral of (12)/((36x^2+1)^2) is arctan(6x)+1/2 sin(2arctan(6x))+C