What is the integral of 1/((x^2+25)^{5/2)} ?

The integral of 1/((x^2+25)^{5/2)} is (2x^3+75x)/(1875(x^2+25)^{3/2)}+C

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integral of 1/((x^2+25)^{5/2)}

\int \frac{1}{( x ^{2} + 25 ) ^{\frac{5}{2}}} d x

\frac{2 x ^{3} + 75 x}{1875 ( x ^{2} + 25 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} + 25 ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{625 sec ^{3} ( u )} d u

= \int \frac{1}{625 sec ^{3} ( u )} d u

Take the constant out:

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

= \frac{1}{625} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{625} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= \frac{1}{625} \cdot \int 1 - v^{2} 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}{625} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{625} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{625} (sin (arctan (\frac{1}{5} x)) - \frac{sin ^{3} ( arctan ( \frac{1}{5} x ) )}{3})

Simplify

\frac{1}{625} (sin (arctan (\frac{1}{5} x)) - \frac{sin ^{3} ( arctan ( \frac{1}{5} x ) )}{3}) : \frac{2 x ^{3} + 75 x}{1875 ( x ^{2} + 25 ) ^{\frac{3}{2}}}

= \frac{2 x ^{3} + 75 x}{1875 ( x ^{2} + 25 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{2 x ^{3} + 75 x}{1875 ( x ^{2} + 25 ) ^{\frac{3}{2}}} + C

f (t) = e^{t a n (π t)}

y = \frac{6 x ^{2} + 2 x + 6}{x}

\frac{d y}{d x} - 2 x y = 10 x

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

\frac{\partial}{\partial x} (3 x y)

What is the integral of 1/((x^2+25)^{5/2)} ?
The integral of 1/((x^2+25)^{5/2)} is (2x^3+75x)/(1875(x^2+25)^{3/2)}+C