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$$\int_1^e\dfrac{1+x^2\ln x}{x+x^2 \ln x} dx$$

Attempt:

I have tried substitutions like $\ln x = t$, but they are just not helping.

I end up with : $\displaystyle\int_0^1 \dfrac{1+e^{2t}t}{1+ e^t t} dt $

Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?

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  • $\begingroup$ $\ln x=t$ is a good substitution. $\endgroup$
    – Nosrati
    Jul 24 '18 at 2:57
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The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.


We need not use any substitutions. Rather, we can write

$$\begin{align} \int_1^e \frac{1+x^2\log(x)}{x+x^2\log(x)}\,dx&=\int_1^e \frac{1-x+x+x^2\log(x)}{x+x^2\log(x)}\,dx\\\\ &=(e-1)+\int_1^e \frac{1-x}{x+x^2\log(x)}\,dx\\\\ &=(e-1)+\int_1^e \left(\frac{1-x}{x+x^2\log(x)}-\frac1x\right)\,dx+\int_1^e \frac1x\,dx\\\\ &=e-\int_1^e \frac{1+\log(x)}{1+x\log(x)}\,dx\\\\ &=e-\left.\left(\log(1+x\log(x)) \right)\right|_1^e\\\\ &=e-\log(1+e) \end{align}$$

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    $\begingroup$ Hi Mark ! May be, you could be more precise and recall that that $(1+x\log(x))'=1+\log(x)$ $\endgroup$ Jul 24 '18 at 4:37
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Let $$I = \int \frac{1+x^2\ln x}{x+x^2\ln x}dx$$

Divide both Numerator and Denominator by $x^2$

so $$I = \int \frac{\frac{1}{x^2}+\ln x}{\frac{1}{x}+\ln x}dx = \int \frac{\bigg(\frac{1}{x}+\ln x\bigg)-\bigg(\frac{1}{x}-\frac{1}{x^2}\bigg)}{\frac{1}{x}+\ln x}dx$$

so $$I = x-\ln \bigg|\frac{1}{x}+\ln x\bigg|+\mathcal{C}.$$

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  • $\begingroup$ (+1) Nicely done. $\endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22
  • $\begingroup$ @DXT short and sweet(+1) $\endgroup$
    – Rishi
    Jun 1 at 10:45

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