What is the integral of (3x^2+14x)(x^3+7x^2+1) ?

The integral of (3x^2+14x)(x^3+7x^2+1) is (x^6)/2+7x^5+(49x^4)/2+x^3+7x^2+C

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integral of (3x^2+14x)(x^3+7x^2+1)

\int (3 x^{2} + 14 x) (x^{3} + 7 x^{2} + 1) d x

\frac{x ^{6}}{2} + 7 x^{5} + \frac{49 x ^{4}}{2} + x^{3} + 7 x^{2} + C

Solution steps

\int (3 x^{2} + 14 x) (x^{3} + 7 x^{2} + 1) d x

Expand

(3 x^{2} + 14 x) (x^{3} + 7 x^{2} + 1) : 3 x^{5} + 35 x^{4} + 98 x^{3} + 3 x^{2} + 14 x

= \int 3 x^{5} + 35 x^{4} + 98 x^{3} + 3 x^{2} + 14 x 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 3 x^{5} d x + \int 35 x^{4} d x + \int 98 x^{3} d x + \int 3 x^{2} d x + \int 14 x d x

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int 35 x^{4} d x = 7 x^{5}

\int 98 x^{3} d x = \frac{49 x ^{4}}{2}

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

\int 14 x d x = 7 x^{2}

= \frac{x ^{6}}{2} + 7 x^{5} + \frac{49 x ^{4}}{2} + x^{3} + 7 x^{2}

Add a constant to the solution

= \frac{x ^{6}}{2} + 7 x^{5} + \frac{49 x ^{4}}{2} + x^{3} + 7 x^{2} + C

What is the integral of (3x^2+14x)(x^3+7x^2+1) ?
The integral of (3x^2+14x)(x^3+7x^2+1) is (x^6)/2+7x^5+(49x^4)/2+x^3+7x^2+C