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

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

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

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

\frac{1}{3} x^{2} (x^{2} + 1) - \frac{1}{3} (\frac{x ^{4}}{4} - \frac{x ^{2}}{2}) + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{3} x (x^{2} - 1) d x = \frac{1}{3} (\frac{x ^{4}}{4} - \frac{x ^{2}}{2})

= \frac{1}{3} x^{2} (x^{2} + 1) - \frac{1}{3} (\frac{x ^{4}}{4} - \frac{x ^{2}}{2})

Add a constant to the solution

= \frac{1}{3} x^{2} (x^{2} + 1) - \frac{1}{3} (\frac{x ^{4}}{4} - \frac{x ^{2}}{2}) + C

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