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

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

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

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

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

Solution steps

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

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} y)

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

Simplify

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

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

Add a constant to the solution

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

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\int (3 - x^{2}) \cdot e^{x} d x

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