What is the integral of (7x^3-4x^2+11x-e^2+17) ?

The integral of (7x^3-4x^2+11x-e^2+17) is (7x^4)/4-(4x^3)/3+(11x^2)/2-e^2x+17x+C

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integral of (7x^3-4x^2+11x-e^2+17)

\int (7 x^{3} - 4 x^{2} + 11 x - e^{2} + 17) d x

\frac{7 x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{11 x ^{2}}{2} - e^{2} x + 17 x + C

Solution steps

\int 7 x^{3} - 4 x^{2} + 11 x - e^{2} + 17 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 7 x^{3} d x - \int 4 x^{2} d x + \int 11 x d x - \int e^{2} d x + \int 17 d x

\int 7 x^{3} d x = \frac{7 x ^{4}}{4}

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

\int 11 x d x = \frac{11 x ^{2}}{2}

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

\int 17 d x = 17 x

= \frac{7 x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{11 x ^{2}}{2} - e^{2} x + 17 x

Add a constant to the solution

= \frac{7 x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{11 x ^{2}}{2} - e^{2} x + 17 x + C

What is the integral of (7x^3-4x^2+11x-e^2+17) ?
The integral of (7x^3-4x^2+11x-e^2+17) is (7x^4)/4-(4x^3)/3+(11x^2)/2-e^2x+17x+C