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

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

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

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

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

Solution steps

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

4 = 2

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

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

Add a constant to the solution

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

\int \frac{1 - 3 x}{2 x - 3 x ^{2}} d x

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

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

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

\int \frac{1}{t t ^{2} - 4} d t

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