What is the integral of (2x-3)(x^2-3x-1)^4ln(x^2-3x-1) ?

The integral of (2x-3)(x^2-3x-1)^4ln(x^2-3x-1) is 1/5 (x^2-3x-1)^5ln(x^2-3x-1)-((x^2-3x-1)^5)/(25)+C

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integral of (2x-3)(x^2-3x-1)^4ln(x^2-3x-1)

\int (2 x - 3) (x^{2} - 3 x - 1)^{4} ln (x^{2} - 3 x - 1) d x

\frac{1}{5} (x^{2} - 3 x - 1)^{5} ln (x^{2} - 3 x - 1) - \frac{( x ^{2} - 3 x - 1 ) ^{5}}{25} + C

Solution steps

\int (2 x - 3) (x^{2} - 3 x - 1)^{4} ln (x^{2} - 3 x - 1) d x

Apply u-substitution

: \int u^{4} ln (u) d u

= \int u^{4} ln (u) d u

Apply Integration By Parts

: \frac{1}{5} u^{5} ln (u) - \int \frac{u ^{4}}{5} d u

= \frac{1}{5} u^{5} ln (u) - \int \frac{u ^{4}}{5} d u

\int \frac{u ^{4}}{5} d u = \frac{u ^{5}}{25}

= \frac{1}{5} u^{5} ln (u) - \frac{u ^{5}}{25}

Substitute back

u = x^{2} - 3 x - 1

= \frac{1}{5} (x^{2} - 3 x - 1)^{5} ln (x^{2} - 3 x - 1) - \frac{( x ^{2} - 3 x - 1 ) ^{5}}{25}

Add a constant to the solution

= \frac{1}{5} (x^{2} - 3 x - 1)^{5} ln (x^{2} - 3 x - 1) - \frac{( x ^{2} - 3 x - 1 ) ^{5}}{25} + C

\int x e^{y^{2}} d x

\frac{d}{d x} ((x y)^{\frac{1}{2}})

cos (5 x)

\int \frac{1}{2 x + 5} d x

x \to 0 lim (x + 3)

What is the integral of (2x-3)(x^2-3x-1)^4ln(x^2-3x-1) ?
The integral of (2x-3)(x^2-3x-1)^4ln(x^2-3x-1) is 1/5 (x^2-3x-1)^5ln(x^2-3x-1)-((x^2-3x-1)^5)/(25)+C