What is the integral of 9ln(x^2-1) ?

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

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integral of 9ln(x^2-1)

\int 9 ln (x^{2} - 1) d x

9 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

Solution steps

\int 9 ln (x^{2} - 1) d x

Take the constant out:

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

= 9 \cdot \int ln (x^{2} - 1) d x

Apply Integration By Parts

: x ln (x^{2} - 1) - \int \frac{2 x ^{2}}{x ^{2} - 1} d x

= 9 (x ln (x^{2} - 1) - \int \frac{2 x ^{2}}{x ^{2} - 1} d x)

\int \frac{2 x ^{2}}{x ^{2} - 1} d x = - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x

= 9 (x ln (x^{2} - 1) - (- ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + 2 x))

Simplify

= 9 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x)

Add a constant to the solution

= 9 (x ln (x^{2} - 1) + ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 2 x) + C

y^{^{'}} + \frac{y}{x} = \frac{y ^{2}}{x}

\frac{d}{d x} (x^{4} - 5 x^{2} + 4)

y = \frac{4}{( 4 - x ^{4} ) ^{\frac{3}{2}}}

\int_{0}^{\frac{π}{2}} 9 (cos (x))^{2} d x

\frac{d}{d x} (x^{3} + 3 x^{2} - 24 x)

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