What is the integral of 6/((9t^2+1)^2) ?

The integral of 6/((9t^2+1)^2) is arctan(3t)+1/2 sin(2arctan(3t))+C

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integral of 6/((9t^2+1)^2)

\int \frac{6}{( 9 t ^{2} + 1 ) ^{2}} d t

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

Solution steps

\int \frac{6}{( 9 t ^{2} + 1 ) ^{2}} d t

Take the constant out:

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

= 6 \cdot \int \frac{1}{( 9 t ^{2} + 1 ) ^{2}} d t

Apply Integral Substitution

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

= 6 \cdot \frac{1}{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

= 6 \cdot \frac{1}{3} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= 6 \cdot \frac{1}{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

= 6 \cdot \frac{1}{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

= 6 \cdot \frac{1}{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)

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 6 \cdot \frac{1}{3} \cdot \frac{1}{2} (arctan (3 t) + \frac{1}{2} sin (2 arctan (3 t)))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (A x^{2} e^{x})

\int_{0}^{3} 450.268 e^{1.12567 t} d t

x \to \infty lim (e^{x} - x)

\int \frac{x ^{2} + 1}{x ^{4} + 1} d x

2 e^{x} cos (x)

What is the integral of 6/((9t^2+1)^2) ?
The integral of 6/((9t^2+1)^2) is arctan(3t)+1/2 sin(2arctan(3t))+C