What is the integral of ((x^2+1)^3(x+2))/(x^5) ?

The integral of ((x^2+1)^3(x+2))/(x^5) is (x^3)/3+x^2+3x+6ln|x|-3/x-3/(x^2)-1/(3x^3)-1/(2x^4)+C

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integral of ((x^2+1)^3(x+2))/(x^5)

\int \frac{( x ^{2} + 1 ) ^{3} ( x + 2 )}{x ^{5}} d x

\frac{x ^{3}}{3} + x^{2} + 3 x + 6 ln ∣ x ∣ - \frac{3}{x} - \frac{3}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{1}{2 x ^{4}} + C

Solution steps

\int \frac{( x ^{2} + 1 ) ^{3} ( x + 2 )}{x ^{5}} d x

Long division

\frac{( x ^{2} + 1 ) ^{3} ( x + 2 )}{x ^{5}} : x^{2} + 2 x + 3 + \frac{6 x ^{4} + 3 x ^{3} + 6 x ^{2} + x + 2}{x ^{5}}

= \int x^{2} + 2 x + 3 + \frac{6 x ^{4} + 3 x ^{3} + 6 x ^{2} + x + 2}{x ^{5}} 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} d x + \int 2 x d x + \int 3 d x + \int \frac{6 x ^{4} + 3 x ^{3} + 6 x ^{2} + x + 2}{x ^{5}} d x

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

\int 2 x d x = x^{2}

\int 3 d x = 3 x

\int \frac{6 x ^{4} + 3 x ^{3} + 6 x ^{2} + x + 2}{x ^{5}} d x = 6 ln ∣ x ∣ - \frac{3}{x} - \frac{3}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{1}{2 x ^{4}}

= \frac{x ^{3}}{3} + x^{2} + 3 x + 6 ln ∣ x ∣ - \frac{3}{x} - \frac{3}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{1}{2 x ^{4}}

Add a constant to the solution

= \frac{x ^{3}}{3} + x^{2} + 3 x + 6 ln ∣ x ∣ - \frac{3}{x} - \frac{3}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{1}{2 x ^{4}} + C

\int (e^{5 - 2 t}) d t

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What is the integral of ((x^2+1)^3(x+2))/(x^5) ?
The integral of ((x^2+1)^3(x+2))/(x^5) is (x^3)/3+x^2+3x+6ln|x|-3/x-3/(x^2)-1/(3x^3)-1/(2x^4)+C