What is the integral of (x^2)/((x^2+16)^{3/2)} ?

The integral of (x^2)/((x^2+16)^{3/2)} is ln(1/4 |x+(16+x^2)^{1/2}|)-x/((16+x^2)^{1/2)}+C

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integral of (x^2)/((x^2+16)^{3/2)}

\int \frac{x ^{2}}{( x ^{2} + 16 ) ^{\frac{3}{2}}} d x

ln (\frac{1}{4} x + (16 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 16 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{2}}{( x ^{2} + 16 ) ^{\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}{4} x)

= ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) - sin (arctan (\frac{1}{4} x))

Simplify

ln tan (arctan (\frac{1}{4} x)) + sec (arctan (\frac{1}{4} x)) - sin (arctan (\frac{1}{4} x)) : ln (\frac{1}{4} x + (16 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 16 + x ^{2} ) ^{\frac{1}{2}}}

= ln (\frac{1}{4} x + (16 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 16 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= ln (\frac{1}{4} x + (16 + x^{2})^{\frac{1}{2}}) - \frac{x}{( 16 + x ^{2} ) ^{\frac{1}{2}}} + C

\frac{\partial}{\partial x} (\frac{6 x}{y ^{2}})

\frac{1}{10 \cdot s}

\int \frac{y}{3} d y

\frac{2}{s ^{2}} \cdot e^{- s}

What is the integral of (x^2)/((x^2+16)^{3/2)} ?
The integral of (x^2)/((x^2+16)^{3/2)} is ln(1/4 |x+(16+x^2)^{1/2}|)-x/((16+x^2)^{1/2)}+C