What is the integral of (x^2e^{ax}+1/(ax)) ?

The integral of (x^2e^{ax}+1/(ax)) is e^{ax} 1/a x^2-2/a (e^{ax} 1/a x-e^{ax} 1/(a^2))+1/a ln|x|+C

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integral of (x^2e^{ax}+1/(ax))

\int (x^{2} e^{a x} + \frac{1}{a x}) d x

e^{a x} \frac{1}{a} x^{2} - \frac{2}{a} (e^{a x} \frac{1}{a} x - e^{a x} \frac{1}{a ^{2}}) + \frac{1}{a} ln ∣ x ∣ + C

Solution steps

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

\int x^{2} e^{a x} d x = e^{a x} \frac{1}{a} x^{2} - \frac{2}{a} (e^{a x} \frac{1}{a} x - e^{a x} \frac{1}{a ^{2}})

\int \frac{1}{a x} d x = \frac{1}{a} ln ∣ x ∣

= e^{a x} \frac{1}{a} x^{2} - \frac{2}{a} (e^{a x} \frac{1}{a} x - e^{a x} \frac{1}{a ^{2}}) + \frac{1}{a} ln ∣ x ∣

Add a constant to the solution

= e^{a x} \frac{1}{a} x^{2} - \frac{2}{a} (e^{a x} \frac{1}{a} x - e^{a x} \frac{1}{a ^{2}}) + \frac{1}{a} ln ∣ x ∣ + C

y^{^{''}} - 12 y^{^{'}} + 35 y = 0

\int cos^{4} (x) tan^{3} (x) d x

\frac{1}{x ^{2}}, x, 8 x

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

x \to 0 lim (\frac{e ^{x} cos ( x )}{4})

What is the integral of (x^2e^{ax}+1/(ax)) ?
The integral of (x^2e^{ax}+1/(ax)) is e^{ax} 1/a x^2-2/a (e^{ax} 1/a x-e^{ax} 1/(a^2))+1/a ln|x|+C