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

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

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

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

ln (\frac{1}{3} x + (9 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 9 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

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

Apply Trigonometric Substitution

= \int tan^{2} (u) cos (u) d u

Simplify

tan^{2} (u) cos (u) : sin (u) tan (u)

= \int sin (u) tan (u) d u

Rewrite using trig identities

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

Expand

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

= \int \frac{1}{cos ( u )} - cos (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

= \int \frac{1}{cos ( u )} d u - \int cos (u) d u

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

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

= ln ∣ tan (u) + sec (u) ∣ - sin (u)

Substitute back

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

= ln tan (arctan (\frac{1}{3} x)) + sec (arctan (\frac{1}{3} x)) - sin (arctan (\frac{1}{3} x))

Simplify

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

= ln (\frac{1}{3} x + (9 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 9 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

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

\int sec^{3} (β) d β

\int (2 x + 7) d x

\int \frac{y}{x ^{5} + 1} d y

\int (\frac{x + 1}{x}) d x

\int (\frac{y}{2 x + 3 y}) d x

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