What is the integral of (3x^3-5x^2+3x+4) ?

The integral of (3x^3-5x^2+3x+4) is (3x^4)/4-(5x^3)/3+(3x^2)/2+4x+C

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integral of (3x^3-5x^2+3x+4)

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

\frac{3 x ^{4}}{4} - \frac{5 x ^{3}}{3} + \frac{3 x ^{2}}{2} + 4 x + C

Solution steps

\int 3 x^{3} - 5 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 3 x^{3} d x - \int 5 x^{2} d x + \int 3 x d x + \int 4 d x

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

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

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

\int 4 d x = 4 x

= \frac{3 x ^{4}}{4} - \frac{5 x ^{3}}{3} + \frac{3 x ^{2}}{2} + 4 x

Add a constant to the solution

= \frac{3 x ^{4}}{4} - \frac{5 x ^{3}}{3} + \frac{3 x ^{2}}{2} + 4 x + C

f (x) = 1 + 2 x + x^{2}, at x = - 1

\int (sec (x))^{2} d x

x \to 0 + lim (\frac{5}{sin ( x )})

\int e^{7 x + e^{7 x}} d x

f (x) = 4 x^{3} + 3 x^{2} - 2

What is the integral of (3x^3-5x^2+3x+4) ?
The integral of (3x^3-5x^2+3x+4) is (3x^4)/4-(5x^3)/3+(3x^2)/2+4x+C