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

The integral of (ax+b+x^2)^2 is (ax^4)/2+(2bx^3)/3+(x^5)/5+(a^2x^3)/3+abx^2+b^2x+C

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integral of (ax+b+x^2)^2

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

\frac{a x ^{4}}{2} + \frac{2 b x ^{3}}{3} + \frac{x ^{5}}{5} + \frac{a ^{2} x ^{3}}{3} + ab x^{2} + b^{2} x + C

Solution steps

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

Expand

(a x + b + x^{2})^{2} : 2 a x^{3} + 2 b x^{2} + x^{4} + a^{2} x^{2} + 2 ab x + b^{2}

= \int 2 a x^{3} + 2 b x^{2} + x^{4} + a^{2} x^{2} + 2 ab x + b^{2} 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 2 b x^{2} d x + \int x^{4} d x + \int a^{2} x^{2} d x + \int 2 ab x d x + \int b^{2} d x

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

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

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

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

\int 2 ab x d x = ab x^{2}

\int b^{2} d x = b^{2} x

= \frac{a x ^{4}}{2} + \frac{2 b x ^{3}}{3} + \frac{x ^{5}}{5} + \frac{a ^{2} x ^{3}}{3} + ab x^{2} + b^{2} x

Add a constant to the solution

= \frac{a x ^{4}}{2} + \frac{2 b x ^{3}}{3} + \frac{x ^{5}}{5} + \frac{a ^{2} x ^{3}}{3} + ab x^{2} + b^{2} x + C

3 y^{3} d x - 6 x^{4} d y = 0

\frac{d}{d x} (\frac{7 - sec ( x )}{tan ( x )})

\frac{d}{d x} ((- x^{2} + 4 x + 5) (4 x^{4} - 3))

\frac{d x}{d t} = 3 x t^{3}

\frac{d y}{d x} - \frac{2 y}{x} = x^{2} - 1

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