What is the integral of x(b-x)^2 ?

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

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integral of x(b-x)^2

\int x (b - x)^{2} d x

\frac{1}{2} b^{2} x^{2} - b x^{3} + \frac{1}{2} x^{4} + \frac{b x ^{3}}{3} - \frac{x ^{4}}{4} + C

Solution steps

\int x (b - x)^{2} d x

Apply Integration By Parts

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

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

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

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

Simplify

\frac{1}{2} x^{2} (b - x)^{2} - (- \frac{b x ^{3}}{3} + \frac{x ^{4}}{4}) : \frac{1}{2} b^{2} x^{2} - b x^{3} + \frac{1}{2} x^{4} + \frac{b x ^{3}}{3} - \frac{x ^{4}}{4}

= \frac{1}{2} b^{2} x^{2} - b x^{3} + \frac{1}{2} x^{4} + \frac{b x ^{3}}{3} - \frac{x ^{4}}{4}

Add a constant to the solution

= \frac{1}{2} b^{2} x^{2} - b x^{3} + \frac{1}{2} x^{4} + \frac{b x ^{3}}{3} - \frac{x ^{4}}{4} + C

\int \frac{( ln ( x ) ) ^{\frac{3}{2}}}{x} d x

\int \frac{y}{3 + y ^{2}} d y

\int \frac{1}{w} d w

\int \frac{1}{u ^{2} - a ^{2}} d u

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