What is the integral of (x^2)/((x^2+8)^{3/2)} ?

The integral of (x^2)/((x^2+8)^{3/2)} is ln|x+(8+x^2)^{1/2}|-3/2 ln(2)-x/((8+x^2)^{1/2)}+C

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integral of (x^2)/((x^2+8)^{3/2)}

\int \frac{x ^{2}}{( x ^{2} + 8 ) ^{\frac{3}{2}}} d x

ln x + (8 + x^{2})^{\frac{1}{2}} - \frac{3}{2} ln (2) - \frac{x}{( 8 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{2}}{( x ^{2} + 8 ) ^{\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}{2 \cdot 2 ^{\frac{1}{2}}} x)

= ln tan (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) - sin (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x))

Simplify

ln tan (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) + sec (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) - sin (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) : ln x + (8 + x^{2})^{\frac{1}{2}} - \frac{3}{2} ln (2) - \frac{x}{( 8 + x ^{2} ) ^{\frac{1}{2}}}

= ln x + (8 + x^{2})^{\frac{1}{2}} - \frac{3}{2} ln (2) - \frac{x}{( 8 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= ln x + (8 + x^{2})^{\frac{1}{2}} - \frac{3}{2} ln (2) - \frac{x}{( 8 + x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d r}{d θ} = π

x^{3} - 8 x

\int 18 x^{\frac{1}{2}} d x

\frac{\partial}{\partial x} (- \frac{1}{( x - y ) ^{2}})

f (y) = cos (y)

What is the integral of (x^2)/((x^2+8)^{3/2)} ?
The integral of (x^2)/((x^2+8)^{3/2)} is ln|x+(8+x^2)^{1/2}|-3/2 ln(2)-x/((8+x^2)^{1/2)}+C