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$$\int \frac{1}{x\sqrt{x^2+x+1}}\mathrm{d}x\; ;$$ $$\int\frac{6\sin(2x) + 6 + \ln(x^6)}{x\ln(x) + \sin^2(x)}\mathrm{d}x$$

I have done the first one using Euler's substitution, but I am not sure if it is correct, I have used the substitution for $a>0$, for the second one I have no idea how to start?

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    $\begingroup$ "I am not sure if it is correct" please tell us what answer you got, otherwise we can't verify. Not everybody is clear with Euler's substitution so mention it. Are these in a textbook? Which chapter is it part of? All these questions need to be addressed in the question post, because the best way to be instructive is to solve the question with the techniques that you know or are supposed to use, so you need to tell us that. $\endgroup$ – Teresa Lisbon Jan 19 at 14:05
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    $\begingroup$ please try not to use images when their content is essential in understanding the question. $\endgroup$ – Thomas Prévost Jan 19 at 14:06
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    $\begingroup$ For the second one, you need to notice that the numerator is six times the derivative of the denominator. $\endgroup$ – Tavish Jan 19 at 14:06
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    $\begingroup$ And for the first one, you might find x=(1/z) substitution useful $\endgroup$ – ConfusedHuman Jan 19 at 14:09
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$$I=\int \frac{1}{x\sqrt{x^2+x+1}}\mathrm{d}x$$ Let $x=\dfrac{1}{t} \implies dx=-\dfrac{dt}{t^2}$ $$I=-\int \frac{dt}{\sqrt{t^2+t+1}}=-\int \frac{dt}{\sqrt{\left(t+\dfrac12 \right)^2+\dfrac34}} $$ $$=-\ln\left[t+\frac12+\sqrt{t^2+t+1}\right]+C, ~ \text{where} ~t=1/x$$

$$J=\int\frac{6\sin(2x) + 6 + \ln(x^6)}{x\ln(x) + \sin^2(x)}\mathrm{d}x$$ Let $$x\ln x+\sin^2 x=t \implies (\ln x+1+\sin 2x)dx=dt$$ $$\implies J=\int \frac{6dt}{t}=6\ln (x\ln x+\sin x)+C$$

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    $\begingroup$ Thanks I corrected it now. $\endgroup$ – Z Ahmed Jan 19 at 15:05
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Per the Euler substitution $\sqrt{x^2+x+1}=x+t$ $$\int \frac{1}{x\sqrt{x^2+x+1}}{d}x=2\int \frac1{t^2-1}dt=\ln\frac{t-1}{t+1}+C $$

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