What is the integral from 1 to t of xln(x) ?

The integral from 1 to t of xln(x) is (t^2ln(t))/2+(-t^2+1)/4

Popular >

integral from 1 to t of xln(x)

\int_{1}^{t} x ln (x) d x

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

Solution steps

\int_{1}^{t} x ln (x) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} ln (x) - \int \frac{x}{2} d x]_{1}^{t}

= [\frac{1}{2} x^{2} ln (x) - \int \frac{x}{2} d x]_{1}^{t}

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

= [\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}]_{1}^{t}

Compute the boundaries:

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

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

Simplify

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

\int_{- 5}^{0} 2 + 4 - x + x d x

\int_{1}^{4} \frac{lo g _{2} ( x )}{x} d x

\int_{1}^{4} x + 6 d x

\int_{1}^{e} (24 + ln (x))^{2} d x

\int_{2}^{3} \frac{3}{1 + x ^{2}} d x

What is the integral from 1 to t of xln(x) ?
The integral from 1 to t of xln(x) is (t^2ln(t))/2+(-t^2+1)/4