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

The integral of (x^2+3x-2)^2 is 1/160 (2x+3)^5-17/48 (2x+3)^3+289/16 (x+3/2)+C

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

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

\frac{1}{160} (2 x + 3)^{5} - \frac{17}{48} (2 x + 3)^{3} + \frac{289}{16} (x + \frac{3}{2}) + C

Solution steps

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

Complete the square

x^{2} + 3 x - 2 : (x + \frac{3}{2})^{2} - \frac{17}{4}

= \int ((x + \frac{3}{2})^{2} - \frac{17}{4})^{2} d x

Apply u-substitution

: \int (u^{2} - \frac{17}{4})^{2} d u

= \int (u^{2} - \frac{17}{4})^{2} d u

Expand

(u^{2} - \frac{17}{4})^{2} : u^{4} - \frac{17 u ^{2}}{2} + \frac{289}{16}

= \int u^{4} - \frac{17 u ^{2}}{2} + \frac{289}{16} d u

Apply the Sum Rule:

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

= \int u^{4} d u - \int \frac{17 u ^{2}}{2} d u + \int \frac{289}{16} d u

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

\int \frac{17 u ^{2}}{2} d u = \frac{17 u ^{3}}{6}

\int \frac{289}{16} d u = \frac{289}{16} u

= \frac{u ^{5}}{5} - \frac{17 u ^{3}}{6} + \frac{289}{16} u

Substitute back

u = x + \frac{3}{2}

= \frac{( x + \frac{3}{2} ) ^{5}}{5} - \frac{17 ( x + \frac{3}{2} ) ^{3}}{6} + \frac{289}{16} (x + \frac{3}{2})

Simplify

\frac{( x + \frac{3}{2} ) ^{5}}{5} - \frac{17 ( x + \frac{3}{2} ) ^{3}}{6} + \frac{289}{16} (x + \frac{3}{2}) : \frac{1}{160} (2 x + 3)^{5} - \frac{17}{48} (2 x + 3)^{3} + \frac{289}{16} (x + \frac{3}{2})

= \frac{1}{160} (2 x + 3)^{5} - \frac{17}{48} (2 x + 3)^{3} + \frac{289}{16} (x + \frac{3}{2})

Add a constant to the solution

= \frac{1}{160} (2 x + 3)^{5} - \frac{17}{48} (2 x + 3)^{3} + \frac{289}{16} (x + \frac{3}{2}) + C

g (x) = y = 2 x, x = 1

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What is the integral of (x^2+3x-2)^2 ?
The integral of (x^2+3x-2)^2 is 1/160 (2x+3)^5-17/48 (2x+3)^3+289/16 (x+3/2)+C