Evaluate the integral. ∫012ln⁡(x)xdx

Question

Evaluate the integral.  ∫012ln⁡(x)xdx

turtletalk75 · Accepted Answer

Step 1  Consider the provided integral,  ∫012ln⁡(x)xdx  Evaluate the above integral.  Step 2  Now, solve the above integral using by part formula.  ∫012ln⁡(x)xdx=2⋅∫01ln⁡(x)xdx  =2[2xln⁡(x)−∫2xdx]01  =2[2xln⁡(x)−4x]01  =2(-4)  =-8

sirpsta3u · Accepted Answer

∫2ln⁡(x)xdx  Lets

nick1337 · Accepted Answer

The integration process can be simplified by making a change of variables:
$t = \sqrt{x}, x = t^{2}, d x = 2 * t * d t$
Then the original integral can be written as follows: The
$\int 4 * \ln (t^{2}) * d t$
$\int 4 * \ln (x^{2}) * d x$
formula for integration by parts:
$\int U (x) * d V (x) = U (x) * V (x) - \int V (x) * d U (x)$
Put
$U = 4 \ln (x^{2})$
dV=dx
Then:
$d U = \frac{8}{x} * d x$
V=x
Therefore:
$\int 4 * \ln (x^{2}) * d x = 4 * x * \ln (x^{2}) - \int 8 * d x$
Find the integral
$\int 8 * d x = 8 * x$
Answer:
$\int 4 * \ln (x^{2}) = 4 * x * \ln (x^{2}) - 8 * x + C$
or
$\int 4 * \ln (x^{2}) = 4 * x * (\ln (x^{2}) - 2) + C$
To write the final answer, it remains to substitute sqrt (x) instead of t.
$4 \sqrt{x} (\ln (x) - 2) + C$
Let's calculate a definite integral:
$\int_{0}^{1} \frac{2 * \ln (x)}{\sqrt{x}} * d x = (4 \sqrt{x} * (\ln (x) - 2)) |_{0}^{1}$
F(1)=-8
F(0)=
I=-8-(0)=-8

Evaluate the integral. \int_{0}^{1}\frac{2\ln (x)}{\sqrt{x}}dx

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