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

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

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

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

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

Solution steps

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

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

x \to 2 lim (3 x^{2} + 4 x + 10)

\int_{π}^{2 π} cos (x) d x

4 cot (\frac{x}{7})

\frac{d}{d x} ((8 (2 x)^{4} + 5)^{3})

2 x^{3} - 3 x^{2} + 5

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