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Question Number 61240 by Tawa1 last updated on 30/May/19

$$\int \frac{{\mathrm{x}}^2-4}{{({\mathrm{x}}^2+4)}^2}\mathrm{dx}$$

Commented by maxmathsup by imad last updated on 31/May/19

$$letA=\int \frac{x^2-4}{{(x^2+4)}^2}dx\Rightarrow A=\int \frac{x^2+4-8}{{(x^2+4)}^2}dx$$$$=\int \frac{dx}{x^2+4}-8\int \frac{dx}{{(x^2+1)}^2}$$$$\int \frac{dx}{x^2+4}{=}_{x=2t}\int \frac{2dt}{4(1+t^2)}=\frac{1}{2}arctan\left(t\right)+c_0$$$$=\frac{1}{2}arctan\left(\frac{x}{2}\right)+c_0$$$$\int \frac{dx}{{(x^2+1)}^2}{=}_{x=tan\theta }\int \frac{(1+{tan}^2\theta )}{{(1+{tan}^2\theta )}^2}d\theta =\int \frac{d\theta }{1+{tan}^2\theta }$$$$=\int {cos}^2\theta d\theta =\frac{1}{2}\int (1+cos\left(2\theta \right))d\theta$$$$=\frac{1}{2}\theta +\frac{1}{4}sin\left(2\theta \right)+c_1=\frac{\theta }{2}+\frac{1}{2}sin\theta cos\theta$$$$=\frac{1}{2}arctan\left(x\right)+\frac{1}{2}\frac{x}{\sqrt{1+x^2}}\frac{1}{\sqrt{1+x^2}}=\frac{1}{2}arctan\left(x\right)+\frac{x}{2(1+x^2)}+c_1\Rightarrow$$$$A=\frac{1}{2}arctan\left(\frac{x}{2}\right)-4arctan\left(x\right)-\frac{4x}{1+x^2}+C.$$

Commented by Tawa1 last updated on 01/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 01/Jun/19

$$youarewelcome.$$

Answered by MJS last updated on 30/May/19

$$\mathrm{in}\mathrm{such}\mathrm{cases}\mathrm{I}\mathrm{try}\mathrm{the}\mathrm{following}:$$$$\delta \left(x\right)=\frac{d}{dx}\left[\frac{f\left(x\right)}{x^2+4}\right]=\frac{f^{\prime }\left(x\right)(x^2+4)-2f\left(x\right)x}{{(x^2+4)}^2}$$$$\mathrm{if}f\left(x\right)=c\Rightarrow \delta \left(x\right)=-\frac{2cx}{{(x^2+4)}^2}$$$$\Rightarrow \mathrm{we}\mathrm{need}\mathrm{another}x$$$$\mathrm{if}f\left(x\right)=cx\Rightarrow \delta \left(x\right)=\frac{-c(x^2-4)}{{(x^2+4)}^2}$$$$\Rightarrow c=-1\Rightarrow f\left(x\right)=-x\Rightarrow$$$$\int \frac{x^2-4}{{(x^2+4)}^2}dx=-\frac{x}{x^2+4}$$

Commented by Tawa1 last updated on 01/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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