What is the integral of xln(sqrt(x)) ?

The integral of xln(sqrt(x)) is 1/4 x^2ln(x)-(x^2)/8+C

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integral of xln(sqrt(x))

\int x ln (x) d x

\frac{1}{4} x^{2} ln (x) - \frac{x ^{2}}{8} + C

Solution steps

\int x ln (x) d x

Apply Integration By Parts

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

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

\int \frac{x}{4} d x = \frac{x ^{2}}{8}

= \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{8}

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

= \frac{1}{4} x^{2} ln (x) - \frac{x ^{2}}{8}

Add a constant to the solution

= \frac{1}{4} x^{2} ln (x) - \frac{x ^{2}}{8} + C

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

\int (tan (x))^{5} d x

\frac{d}{d x} (ln (1 + x^{3}))

y = x^{3} - 2 x + 1, (0, 1)

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

What is the integral of xln(sqrt(x)) ?
The integral of xln(sqrt(x)) is 1/4 x^2ln(x)-(x^2)/8+C