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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

\int \frac{x ^{2} - 2}{x ^{2} - x - 2} d x

\int sec^{- 2} (x) tan^{3} (x) d x

\int \frac{1}{49 - 25 x ^{2}} d x

\int (x - x^{3} - 2) x d x

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