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

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

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integral of (x^2)/((sqrt(x^2+9))^3)

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

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

Solution steps

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

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

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

Apply Trigonometric Substitution

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

= \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{x}{9 + x ^{2}}

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

Add a constant to the solution

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

x \to 4 lim (x)

x (\frac{d y}{d x}) + 4 y = x^{3} - x

f (x) = 2 x^{2} - 7 x + 5, at x = 1

y = 5 x^{3} - 2 x^{4}, x = 0, x = 2

\frac{d}{d x} (6 x^{2} \cdot cos (2 x))

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