Integral of ln(t) dx

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$$\int\limits_{0}^{1} \log{\left(t \right)}\, dt$$

Integral(log(t), (t, 0, 1))

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(t \right)} = \log{\left(t \right)}$ and let $\operatorname{dv}{\left(t \right)} = 1$ .

Then $\operatorname{du}{\left(t \right)} = \frac{1}{t}$ .

To find $v{\left(t \right)}$ :
1. The integral of a constant is the constant times the variable of integration:
  
  $\int 1\, dt = t$
Now evaluate the sub-integral.
The integral of a constant is the constant times the variable of integration:

$\int 1\, dt = t$
Now simplify:

$t \left(\log{\left(t \right)} - 1\right)$
Add the constant of integration:

$t \left(\log{\left(t \right)} - 1\right)+ \mathrm{constant}$

The answer is:

$t \left(\log{\left(t \right)} - 1\right)+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                            
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 | log(t) dt = C - t + t*log(t)
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$$t\,\log t-t$$

Simplify

The answer [src]

-1

$$-1$$

-1

$$-1$$

Упростить

Numerical answer [src]

-1.0

-1.0

Use the examples entering the upper and lower limits of integration.