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Question Number 86897 by ram roop sharma last updated on 01/Apr/20

If ∫ (1/((x^2 +1)(x^2 +4))) dx = A tan^(−1) x+B tan^(−1) (x/2)+C, then

$$\mathrm{If}\:\int\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:=\:\:{A}\:\mathrm{tan}^{−\mathrm{1}} {x}+{B}\:\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+{C},\:\mathrm{then} \\ $$

Commented by Tony Lin last updated on 01/Apr/20

(d/dx)(Atan^(−1) x+Btan^(−1) (x/2)+C) =(A/(x^2 +1))+(B/(((x/2))^2 +1))×(1/2) =(A/(x^2 +1))+((2B)/(x^2 +4)) =((A(x^2 +4)+2B(x^2 +1))/((x^2 +1)(x^2 +4))) ⇒ { ((A+2B=0)),((4A+2B=1)) :} ⇒A=(1/3) , B=−(1/6)

$$\frac{{d}}{{dx}}\left({Atan}^{−\mathrm{1}} {x}+{Btan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+{C}\right) \\ $$$$=\frac{{A}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{B}}{\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{{A}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}{B}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$=\frac{{A}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\mathrm{2}{B}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$\Rightarrow\begin{cases}{{A}+\mathrm{2}{B}=\mathrm{0}}\\{\mathrm{4}{A}+\mathrm{2}{B}=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{3}}\:,\:{B}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by redmiiuser last updated on 01/Apr/20

A=(1/3) B=−(1/6)

$${A}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$

Commented by redmiiuser last updated on 01/Apr/20

simple way to solve is by partial fractions.

$${simple}\:{way}\:{to}\:{solve} \\ $$$${is}\:{by}\:{partial}\:{fractions}. \\ $$

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