Matematické Fórum

stuart clark · 22. 11. 2013 05:01

$\displaystyle{\int \frac{x^2+2\,x+1+\left(3\,x+1\right)\,\sqrt{x+\ln x}}{x\,\sqrt{x+\ln x}\,\left(x+\sqrt{x+\ln x}\right)}\,dx}$

jardofpr · 08. 12. 2013 14:40

hi ↑ stuart clark:

Autor skryl část textu. Zobrazit/skrýt!

stuart clark · 10. 12. 2013 05:00

Thanks ↑ jardofpr:

Using yours Hint.

$\int\frac{x^2+2x+1+(3x+1)\sqrt{x+\ln (x)}}{x\sqrt{x+\ln(x)}\left(x+\sqrt{x+\ln (x)}\right)}dx$

$= \int\frac{2x\sqrt{x+\ln (x)}+(x+1)+(x+1)\sqrt{x+\ln (x)}+x^2+x}{x\sqrt{x+\ln(x)}\left(x+\sqrt{x+\ln (x)}\right)}dx$

$= \int\frac{2x\sqrt{x+\ln (x)}+(x+1)+(x+1)\left(x+\sqrt{x+\ln(x)}\right)}{x\sqrt{x+\ln(x)}\left(x+\sqrt{x+\ln (x)}\right)}dx$

$= 2\int\frac{2x\sqrt{x+\ln (x)}+(x+1)}{x\sqrt{x+\ln(x)}\left(x+\sqrt{x+\ln (x)}\right)}dx + \int\frac{x+1}{x\sqrt{x+\ln (x)}}dx$

Now Let $\left(x+\sqrt{x+\ln (x)}\right) = t$ , Then $\left(2x\sqrt{x+\ln(x)}+(x+1)\right)dx = dt$ and $\left(x+\ln (x)\right) = u^2$ and $\left(1+\frac{1}{x}\right)dx = 2udu$

So Integral $= 2\left(x+\sqrt{x+\ln (x)}\right)+2\sqrt{x+\ln (x)}+\mathbb{C}$

Matematické Fórum

#1 22. 11. 2013 05:01

Indefinite Integral

#2 08. 12. 2013 14:40

Re: Indefinite Integral

#3 10. 12. 2013 05:00

Re: Indefinite Integral

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