What is the integral of (4x^3+6x^2+3) ?

The integral of (4x^3+6x^2+3) is x^4+2x^3+3x+C

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integral of (4x^3+6x^2+3)

\int (4 x^{3} + 6 x^{2} + 3) d x

x^{4} + 2 x^{3} + 3 x + C

Solution steps

\int 4 x^{3} + 6 x^{2} + 3 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^{3} d x + \int 6 x^{2} d x + \int 3 d x

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

\int 6 x^{2} d x = 2 x^{3}

\int 3 d x = 3 x

= x^{4} + 2 x^{3} + 3 x

Add a constant to the solution

= x^{4} + 2 x^{3} + 3 x + C

\int \frac{3}{3 + x} d x

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

\int x + cos^{2} (x) d x

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

What is the integral of (4x^3+6x^2+3) ?
The integral of (4x^3+6x^2+3) is x^4+2x^3+3x+C