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

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

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

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

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

Solution steps

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Expand

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

= \int x ln (x) + \frac{ln ( x )}{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 ln (x) d x + \int \frac{ln ( x )}{x} d x

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

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

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

Add a constant to the solution

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

\frac{3}{2 x + 4} + 9 x^{2}

x \to 0 + lim ((1 - 2 x)^{\frac{3}{x}})

\int \frac{x ( x + 1 )}{4 x ^{4}} d x

\frac{\partial}{\partial x} (y ln (x + y))

\int 16 e^{- 16 x} d x

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