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can anyone please help to integrate

$$\int_{1}^{3}(x-1)\ln{x}\mathrm dx$$

$u=x-1$

$du=dx$

$dv=\ln{x}$

$v=\int\ln{x}dx$

$$\int_{1}^{3}(x-1)\ln{x}\mathrm dx=(x-1)\int\ln{x}-\int \int\ln{x}dx$$

need help

  • 0
    It might be better to get rid of the natural log by taking the derivative of that instead. – 2017-02-25
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    is it possible to integrate by letting dv=lnx – 2017-02-25

5 Answers 5

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You can use the sustitution $x=e^t$ \begin{align*} \int_1^3(x-1)\ln x\,dx&=\int_0^{\ln3}(e^t-1)te^t\,dt\\ &=\int_1^{\ln 3}(te^{2t}-te^t)\,dt\\ &=\left.\left(\tfrac{t}{2}-\tfrac14\right)e^{2t}-\left(t-1\right)e^t\right|_0^{\ln3} \end{align*}

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Hint:

The integral can be addressed by parts (integrating $(x-1)$). We can work it out differently by "trying" a few derivatives.

$$(x\log x)'=\log x+\frac xx=\log x+1,$$

$$(x^2\log x)'=2x\log x+\frac{x^2}x=2x\log x+x.$$

Then we can conclude that

$$(x\log x-x)'=\log x,$$

$$\left(\frac{x^2}2\log x-\frac{x^2}4\right)'=x\log x.$$

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$\textbf{Hint}$. Try $u=(x-1)\ln x$ and $dv=dx$. Then,

$$\int (x-1)\ln x dx = x(x-1)\ln x-\int x\left(\ln x +1 -\frac{1}{x}\right)dx= x(x-1)\ln x-\int x\ln x dx +\int xdx -\int dx.$$

Now, try for the first integral $w=x\ln x$ and $dy=dx$ and you got it.

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Let $I = \int_{1}^{3} (x-1)lnxdx$

Instead set $u = lnx$, $v=\int x-1dx = \frac{x^{2}}{2} -x$

Then $I = \left | (\frac{x^{2}}{2} -x)lnx \right |_{1}^{3} -\int_{1}^{3}\frac{1}{x}(\frac{x^{2}}{2}-x)dx$

$I = \frac{3}{2}ln3 - \int_{1}^{3}\frac{x}{2}-1dx$

$I = \frac{3}{2}ln3 - \left | \frac{x^{2}}{4}-x \right |_{1}^{3}$

$I = \frac{3}{2}ln3 - 0 = \frac{3}{2}ln3$

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Recall that

$$\ln(x)=\int_1^x\frac1t\ dt$$

Thus, your integral becomes

$$\int_1^3\int_1^x\frac{x-1}t\ dt\ dx=\int_1^3\int_t^3\frac{x-1}t\ dx\ dt=\int_1^3\frac{3+2t-t^2}{2t}\ dt=\frac32\ln(3)$$