What is the integral of 1/(x(x^2+9)^{3/2)} ?

The integral of 1/(x(x^2+9)^{3/2)} is 1/27 (ln|tan(1/2 arctan(1/3 x))|+3/((9+x^2)^{1/2)})+C

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integral of 1/(x(x^2+9)^{3/2)}

\int \frac{1}{x ( x ^{2} + 9 ) ^{\frac{3}{2}}} d x

\frac{1}{27} (ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + \frac{3}{( 9 + x ^{2} ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{1}{x ( x ^{2} + 9 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \int \frac{1}{27 sec ( u ) tan ( u )} d u

Take the constant out:

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

= \frac{1}{27} \cdot \int \frac{1}{sec ( u ) tan ( u )} d u

Rewrite using trig identities

= \frac{1}{27} \cdot \int \frac{1 - sin ^{2} ( u )}{sin ( u )} d u

Expand

\frac{1 - sin ^{2} ( u )}{sin ( u )} : \frac{1}{sin ( u )} - sin (u)

= \frac{1}{27} \cdot \int \frac{1}{sin ( u )} - sin (u) d u

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}{27} (\int \frac{1}{sin ( u )} d u - \int sin (u) d u)

\int \frac{1}{sin ( u )} d u = ln tan (\frac{u}{2})

\int sin (u) d u = - cos (u)

= \frac{1}{27} (ln tan (\frac{u}{2}) - (- cos (u)))

Substitute back

u = arctan (\frac{1}{3} x)

= \frac{1}{27} (ln tan (\frac{arctan ( \frac{1}{3} x )}{2}) - (- cos (arctan (\frac{1}{3} x))))

Simplify

\frac{1}{27} (ln tan (\frac{arctan ( \frac{1}{3} x )}{2}) - (- cos (arctan (\frac{1}{3} x)))) : \frac{1}{27} (ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + \frac{3}{( 9 + x ^{2} ) ^{\frac{1}{2}}})

= \frac{1}{27} (ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + \frac{3}{( 9 + x ^{2} ) ^{\frac{1}{2}}})

Add a constant to the solution

= \frac{1}{27} (ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + \frac{3}{( 9 + x ^{2} ) ^{\frac{1}{2}}}) + C

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What is the integral of 1/(x(x^2+9)^{3/2)} ?
The integral of 1/(x(x^2+9)^{3/2)} is 1/27 (ln|tan(1/2 arctan(1/3 x))|+3/((9+x^2)^{1/2)})+C