What is the integral of (3y^2+1)ln(y^2-1) ?

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

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integral of (3y^2+1)ln(y^2-1)

\int (3 y^{2} + 1) ln (y^{2} - 1) d y

y^{3} ln (y^{2} - 1) - \frac{2 y ^{3}}{3} - 4 y + y ln (y^{2} - 1) + 2 ln ∣ y + 1 ∣ - 2 ln ∣ y - 1 ∣ + C

Solution steps

\int (3 y^{2} + 1) ln (y^{2} - 1) d y

Expand

(3 y^{2} + 1) ln (y^{2} - 1) : 3 y^{2} ln (y^{2} - 1) + ln (y^{2} - 1)

= \int 3 y^{2} ln (y^{2} - 1) + ln (y^{2} - 1) d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 y^{2} ln (y^{2} - 1) d y + \int ln (y^{2} - 1) d y

\int 3 y^{2} ln (y^{2} - 1) d y = y^{3} ln (y^{2} - 1) - \frac{2 y ^{3}}{3} - 2 y + ln ∣ y + 1 ∣ - ln ∣ y - 1 ∣

\int ln (y^{2} - 1) d y = y ln (y^{2} - 1) + ln ∣ y + 1 ∣ - ln ∣ y - 1 ∣ - 2 y

= y^{3} ln (y^{2} - 1) - \frac{2 y ^{3}}{3} - 2 y + ln ∣ y + 1 ∣ - ln ∣ y - 1 ∣ + y ln (y^{2} - 1) + ln ∣ y + 1 ∣ - ln ∣ y - 1 ∣ - 2 y

Simplify

= y^{3} ln (y^{2} - 1) - \frac{2 y ^{3}}{3} - 4 y + y ln (y^{2} - 1) + 2 ln ∣ y + 1 ∣ - 2 ln ∣ y - 1 ∣

Add a constant to the solution

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

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