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

The integral of 1/((16t^2+9)^2) is 1/216 (arctan(4/3 t)+1/2 sin(2arctan(4/3 t)))+C

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

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

\frac{1}{216} (arctan (\frac{4}{3} t) + \frac{1}{2} sin (2 arctan (\frac{4}{3} t))) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{108 ( 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}{108} \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}{108} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{108} \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}{108} \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}{108} \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}{108} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{108} \cdot \frac{1}{2} (arctan (\frac{4}{3} t) + \frac{1}{2} sin (2 arctan (\frac{4}{3} t)))

Simplify

\frac{1}{108} \cdot \frac{1}{2} (arctan (\frac{4}{3} t) + \frac{1}{2} sin (2 arctan (\frac{4}{3} t))) : \frac{1}{216} (arctan (\frac{4}{3} t) + \frac{1}{2} sin (2 arctan (\frac{4}{3} t)))

= \frac{1}{216} (arctan (\frac{4}{3} t) + \frac{1}{2} sin (2 arctan (\frac{4}{3} t)))

Add a constant to the solution

= \frac{1}{216} (arctan (\frac{4}{3} t) + \frac{1}{2} sin (2 arctan (\frac{4}{3} t))) + C

\int - 2 e^{3 x} + \frac{5}{x} d x

x \to 0 lim (18 + \frac{20}{x ^{3}})

\int \frac{1}{( 4 x + 3 ) ^{2}} d x

\int - x sin (z) d z

\int (2 + x + x^{2} + 3 x^{3}) d x

What is the integral of 1/((16t^2+9)^2) ?
The integral of 1/((16t^2+9)^2) is 1/216 (arctan(4/3 t)+1/2 sin(2arctan(4/3 t)))+C