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

The integral of (x^2+1)(2x-3) is (2x^4)/3-(x^4)/6-x^3+x^2-3x+C

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integral of (x^2+1)(2x-3)

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

\frac{2 x ^{4}}{3} - \frac{x ^{4}}{6} - x^{3} + x^{2} - 3 x + C

Solution steps

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

Apply Integration By Parts

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

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

\int 2 (\frac{x ^{3}}{3} + x) d x = \frac{x ^{4}}{6} + x^{2}

= \frac{2 x ^{4}}{3} + 2 x^{2} - x^{3} - 3 x - (\frac{x ^{4}}{6} + x^{2})

Simplify

= \frac{2 x ^{4}}{3} - \frac{x ^{4}}{6} - x^{3} + x^{2} - 3 x

Add a constant to the solution

= \frac{2 x ^{4}}{3} - \frac{x ^{4}}{6} - x^{3} + x^{2} - 3 x + C

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What is the integral of (x^2+1)(2x-3) ?
The integral of (x^2+1)(2x-3) is (2x^4)/3-(x^4)/6-x^3+x^2-3x+C