What is the integral of 1^2(x^3+4x^2-6) ?

The integral of 1^2(x^3+4x^2-6) is (x^4)/4+(4x^3)/3-6x+C

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integral of 1^2(x^3+4x^2-6)

\int 1^{2} (x^{3} + 4 x^{2} - 6) d x

\frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x + C

Solution steps

\int 1^{2} (x^{3} + 4 x^{2} - 6) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1^{2} \cdot \int x^{3} + 4 x^{2} - 6 d x

Simplify

= 1 \cdot \int x^{3} + 4 x^{2} - 6 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

= 1 \cdot (\int x^{3} d x + \int 4 x^{2} d x - \int 6 d x)

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

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

\int 6 d x = 6 x

= 1 \cdot (\frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x)

Simplify

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 6 x + C

\int - 12 e^{x} d x

\int arccos (\frac{x}{x + 1}) d x

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\int 8 \cdot 2^{- x} d x

\frac{4}{x ^{2}}

What is the integral of 1^2(x^3+4x^2-6) ?
The integral of 1^2(x^3+4x^2-6) is (x^4)/4+(4x^3)/3-6x+C