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

The integral of (x^2+ax-3)^2 is (ax^4)/2+(x^5)/5-2x^3+(a^2x^3)/3-3ax^2+9x+C

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

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

\frac{a x ^{4}}{2} + \frac{x ^{5}}{5} - 2 x^{3} + \frac{a ^{2} x ^{3}}{3} - 3 a x^{2} + 9 x + C

Solution steps

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

Expand

(x^{2} + a x - 3)^{2} : 2 a x^{3} + x^{4} - 6 x^{2} + a^{2} x^{2} - 6 a x + 9

= \int 2 a x^{3} + x^{4} - 6 x^{2} + a^{2} x^{2} - 6 a x + 9 d x

Apply the Sum Rule:

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

= \int 2 a x^{3} d x + \int x^{4} d x - \int 6 x^{2} d x + \int a^{2} x^{2} d x - \int 6 a x d x + \int 9 d x

\int 2 a x^{3} d x = \frac{a x ^{4}}{2}

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

\int 6 x^{2} d x = 2 x^{3}

\int a^{2} x^{2} d x = \frac{a ^{2} x ^{3}}{3}

\int 6 a x d x = 3 a x^{2}

\int 9 d x = 9 x

= \frac{a x ^{4}}{2} + \frac{x ^{5}}{5} - 2 x^{3} + \frac{a ^{2} x ^{3}}{3} - 3 a x^{2} + 9 x

Add a constant to the solution

= \frac{a x ^{4}}{2} + \frac{x ^{5}}{5} - 2 x^{3} + \frac{a ^{2} x ^{3}}{3} - 3 a x^{2} + 9 x + C

\int_{0}^{1} 2 x^{2} - 1 d x

\frac{d ^{2}}{d x ^{2}} ((x + 1) e^{x})

n = 1 \sum \infty sin (πn)

\int \frac{x}{1 + 9 x ^{4}} d x

\frac{d}{d z} (\frac{i}{2 i} ln (\frac{i + i z}{i - i z}))

What is the integral of (x^2+ax-3)^2 ?
The integral of (x^2+ax-3)^2 is (ax^4)/2+(x^5)/5-2x^3+(a^2x^3)/3-3ax^2+9x+C