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

The integral of 1/((a^2+x^2)^{5/2)} is (2x^3+3a^2x)/(3a^4(x^2+a^2)^{3/2)}+C

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

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

\frac{2 x ^{3} + 3 a ^{2} x}{3 a ^{4} ( x ^{2} + a ^{2} ) ^{\frac{3}{2}}} + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{a ^{4} sec ^{3} ( u )} d u

= \int \frac{1}{a ^{4} 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}{a ^{4}} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{a ^{4}} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

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

= \frac{1}{a ^{4}} \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}{a ^{4}} (\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}{a ^{4}} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{a ^{4}} (sin (arctan (\frac{1}{a} x)) - \frac{sin ^{3} ( arctan ( \frac{1}{a} x ) )}{3})

Simplify

\frac{1}{a ^{4}} (sin (arctan (\frac{1}{a} x)) - \frac{sin ^{3} ( arctan ( \frac{1}{a} x ) )}{3}) : \frac{2 x ^{3} + 3 a ^{2} x}{3 a ^{4} ( x ^{2} + a ^{2} ) ^{\frac{3}{2}}}

= \frac{2 x ^{3} + 3 a ^{2} x}{3 a ^{4} ( x ^{2} + a ^{2} ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{2 x ^{3} + 3 a ^{2} x}{3 a ^{4} ( x ^{2} + a ^{2} ) ^{\frac{3}{2}}} + C

\frac{d}{d a} (b lo g_{10} (\frac{y}{x} + \frac{0.6 x}{a}))

y = x^{\frac{4}{3}}, y = 2 x^{\frac{1}{3}}

\int \frac{x}{6} d x

(arctan (3 x))^{2}

\int \frac{1}{x ^{2} - 8 x + 20} d x

What is the integral of 1/((a^2+x^2)^{5/2)} ?
The integral of 1/((a^2+x^2)^{5/2)} is (2x^3+3a^2x)/(3a^4(x^2+a^2)^{3/2)}+C