What is the integral of ((ax)^3+e^{-bx}) ?

The integral of ((ax)^3+e^{-bx}) is (a^3x^4)/4-1/b e^{-bx}+C

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integral of ((ax)^3+e^{-bx})

\int ((a x)^{3} + e^{- b x}) d x

\frac{a ^{3} x ^{4}}{4} - \frac{1}{b} e^{- b x} + C

Solution steps

\int (a x)^{3} + e^{- b x} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= \int a^{3} x^{3} + e^{- b x} 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 a^{3} x^{3} d x + \int e^{- b x} d x

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

\int e^{- b x} d x = - \frac{1}{b} e^{- b x}

= \frac{a ^{3} x ^{4}}{4} - \frac{1}{b} e^{- b x}

Add a constant to the solution

= \frac{a ^{3} x ^{4}}{4} - \frac{1}{b} e^{- b x} + C

\int_{0}^{2} x^{3} (7 - x^{2})^{\frac{1}{2}} d x

\int_{0}^{4} (x^{2} - 4 x + 3) d x

4 t \frac{d y}{d t} - 5 y = t

\int - 2 \cdot 3^{6 x} d x

y = cot^{2} (x)

What is the integral of ((ax)^3+e^{-bx}) ?
The integral of ((ax)^3+e^{-bx}) is (a^3x^4)/4-1/b e^{-bx}+C