What is the integral of (4x^2)/((x^2-9)^{3/2)} ?

The integral of (4x^2)/((x^2-9)^{3/2)} is 4(-x/((x^2-9)^{1/2)}+ln(1/3 |(x^2-9)^{1/2}+x|))+C

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integral of (4x^2)/((x^2-9)^{3/2)}

\int \frac{4 x ^{2}}{( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{4 x ^{2}}{( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x ^{2}}{( x ^{2} - 9 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int sec^{3} (u) cot^{2} (u) d u

= 4 \cdot \int sec^{3} (u) cot^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \int \frac{csc ^{2} ( u )}{cos ( u )} d u

Apply Integration By Parts

: - csc (u) - \int - sec (u) d u

= 4 (- csc (u) - \int - sec (u) d u)

\int - sec (u) d u = - ln ∣ tan (u) + sec (u) ∣

= 4 (- csc (u) - (- ln ∣ tan (u) + sec (u) ∣))

Substitute back

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

= 4 (- csc (arcsec (\frac{1}{3} x)) - (- ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x))))

Simplify

4 (- csc (arcsec (\frac{1}{3} x)) - (- ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)))) : 4 (- \frac{x}{( x ^{2} - 9 ) ^{\frac{1}{2}}} + ln (\frac{1}{3} (x^{2} - 9)^{\frac{1}{2}} + x))

= 4 (- \frac{x}{( x ^{2} - 9 ) ^{\frac{1}{2}}} + ln (\frac{1}{3} (x^{2} - 9)^{\frac{1}{2}} + x))

Add a constant to the solution

= 4 (- \frac{x}{( x ^{2} - 9 ) ^{\frac{1}{2}}} + ln (\frac{1}{3} (x^{2} - 9)^{\frac{1}{2}} + x)) + C

\int \frac{e ^{- x}}{e ^{- 2 x} + 1} d x

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

\int \frac{x ^{2} - 2 x + 3}{x} d x

\int \frac{9}{2 - 3 x} d x

What is the integral of (4x^2)/((x^2-9)^{3/2)} ?
The integral of (4x^2)/((x^2-9)^{3/2)} is 4(-x/((x^2-9)^{1/2)}+ln(1/3 |(x^2-9)^{1/2}+x|))+C