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Question Number 29201 by mondodotto@gmail.com last updated on 05/Feb/18

Answered by A1B1C1D1 last updated on 05/Feb/18

Commented by NECx last updated on 05/Feb/18

$$whichappssolvedlikethis?$$

Commented by A1B1C1D1 last updated on 05/Feb/18

$$\mathrm{Some},\mathrm{such}\mathrm{as}\mathrm{the}\mathrm{Okswthe}$$$$\mathrm{calculator}\mathrm{editor}\mathrm{and}\mathrm{the}\mathrm{integral}\mathrm{app}:-)$$

Commented by NECx last updated on 05/Feb/18

$$thanks$$

Commented by Joel578 last updated on 05/Feb/18

$$\mathrm{WolframAlpha}$$

Answered by ajfour last updated on 05/Feb/18

$$I=\frac{1}{2}\int \frac{x^2\left(2xdx\right)}{(x^2+1)(x^2+2)}$$$$let{x}^2=t$$$$I=\frac{1}{2}\int \frac{tdt}{(t+1)(t+2)}$$$$=\frac{1}{2}[-\int \frac{dt}{t+1}+2\int \frac{dt}{t+2}]$$$$=-\frac{1}{2}\ln \mid t+1\mid +\ln \mid t+2\mid +c$$$$=\frac{1}{2}\ln \left[\frac{{(x^2+2)}^2}{1+x^2}\right]+c.$$

Commented by NECx last updated on 05/Feb/18

$$soniceworking.\mathit{g}\mathit{r}\mathit{e}\mathit{a}\mathit{t}\mathit{j}\mathit{o}\mathit{b}\mathit{s}\mathit{i}\mathit{r}!$$

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