What is the integral of 1/((x^2+25)^2) ?

The integral of 1/((x^2+25)^2) is 1/250 (arctan(x/5)+1/2 sin(2arctan(x/5)))+C

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integral of 1/((x^2+25)^2)

\int \frac{1}{( x ^{2} + 25 ) ^{2}} d x

\frac{1}{250} (arctan (\frac{x}{5}) + \frac{1}{2} sin (2 arctan (\frac{x}{5}))) + C

Solution steps

\int \frac{1}{( x ^{2} + 25 ) ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{125 ( 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}{125} \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}{125} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{125} \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}{125} \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}{125} \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}{125} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{125} \cdot \frac{1}{2} (arctan (\frac{x}{5}) + \frac{1}{2} sin (2 arctan (\frac{x}{5})))

Simplify

\frac{1}{125} \cdot \frac{1}{2} (arctan (\frac{x}{5}) + \frac{1}{2} sin (2 arctan (\frac{x}{5}))) : \frac{1}{250} (arctan (\frac{x}{5}) + \frac{1}{2} sin (2 arctan (\frac{x}{5})))

= \frac{1}{250} (arctan (\frac{x}{5}) + \frac{1}{2} sin (2 arctan (\frac{x}{5})))

Add a constant to the solution

= \frac{1}{250} (arctan (\frac{x}{5}) + \frac{1}{2} sin (2 arctan (\frac{x}{5}))) + C

f (x) = (\frac{9}{8}) x^{\frac{8}{9}}

\frac{\partial}{\partial y} (4 - x^{2} - 2 y^{2})

2 r (s^{2} + 1) d r + (r^{4} + 1) d s = 0

\int_{- \infty}^{15} x e^{\frac{x}{5}} d x

x \to 1 lim (\frac{1}{x ^{2} - x})

What is the integral of 1/((x^2+25)^2) ?
The integral of 1/((x^2+25)^2) is 1/250 (arctan(x/5)+1/2 sin(2arctan(x/5)))+C