What is the integral of (4x^2-7x+1) ?

The integral of (4x^2-7x+1) is (4x^3)/3-(7x^2)/2+x+C

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integral of (4x^2-7x+1)

\int (4 x^{2} - 7 x + 1) d x

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

Solution steps

\int 4 x^{2} - 7 x + 1 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 4 x^{2} d x - \int 7 x d x + \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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What is the integral of (4x^2-7x+1) ?
The integral of (4x^2-7x+1) is (4x^3)/3-(7x^2)/2+x+C