What is the integral of x^3-4/(\sqrt[3]{x)}+9/x+6e^x ?

The integral of x^3-4/(\sqrt[3]{x)}+9/x+6e^x is (x^4)/4-6x^{2/3}+9ln|x|+6e^x+C

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integral of x^3-4/(\sqrt[3]{x)}+9/x+6e^x

\frac{x ^{4}}{4} - 6 x^{\frac{2}{3}} + 9 ln ∣ x ∣ + 6 e^{x} + C

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{9}{x} d x = 9 ln ∣ x ∣

\int 6 e^{x} d x = 6 e^{x}

= \frac{x ^{4}}{4} - 6 x^{\frac{2}{3}} + 9 ln ∣ x ∣ + 6 e^{x}

Add a constant to the solution

= \frac{x ^{4}}{4} - 6 x^{\frac{2}{3}} + 9 ln ∣ x ∣ + 6 e^{x} + C

\frac{1}{( s ^{2} - 9 )}

\frac{d y}{d x} + 2 x^{2} y = 11 x^{2}

\int \frac{e ^{2 x} + 1}{e ^{2 x} - 1} d x

x \to 0 - lim (1 + \frac{1}{x})

\int_{- \infty}^{6} 10 x e^{2 x} d x

What is the integral of x^3-4/(\sqrt[3]{x)}+9/x+6e^x ?
The integral of x^3-4/(\sqrt[3]{x)}+9/x+6e^x is (x^4)/4-6x^{2/3}+9ln|x|+6e^x+C