What is the integral of 1/x+(x^5e^x) ?

The integral of 1/x+(x^5e^x) is ln|x|+x^5e^x-5(x^4e^x-4(x^3e^x-3(x^2e^x-2(e^xx-e^x))))+C

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integral of 1/x+(x^5e^x)

\int \frac{1}{x} + (x^{5} e^{x}) d x

ln ∣ x ∣ + x^{5} e^{x} - 5 (x^{4} e^{x} - 4 (x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x})))) + C

Solution steps

\int \frac{1}{x} + x^{5} e^{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 \frac{1}{x} d x + \int x^{5} e^{x} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + x^{5} e^{x} - 5 (x^{4} e^{x} - 4 (x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x}))))

Add a constant to the solution

= ln ∣ x ∣ + x^{5} e^{x} - 5 (x^{4} e^{x} - 4 (x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x})))) + C

\frac{\partial}{\partial x} (x^{4})

\frac{d}{d x} (- e^{- x} x - e^{- x})

\int 3 x^{2} (x^{3} + 7)^{3} d x

(9 x y^{2} - 3) d x + (9 x^{2} y + 4) d y = 0

d er i v a t i v e f (x) = e^{x + 9} + 9

What is the integral of 1/x+(x^5e^x) ?
The integral of 1/x+(x^5e^x) is ln|x|+x^5e^x-5(x^4e^x-4(x^3e^x-3(x^2e^x-2(e^xx-e^x))))+C