Question Number 213376 by RoseAli last updated on 03/Nov/24
$$\int\frac{{dx}}{\:\sqrt{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }} \\ $$
Commented by Frix last updated on 03/Nov/24
![Sometimes just use your brain & experience (d/dx)[((g(x))/( (√(4x^2 +1))))]=((g′(x)(4x^2 +1)−4g(x))/((4x^2 +1)^(3/2) )) g′(x)(4x^2 +1)−4g(x)=1 ⇒ g(x)=x ∫(dx/( (√((4x^2 +1)^3 ))))=(x/( (√(4x^2 +1))))+C](https://www.tinkutara.com/question/Q213377.png)
$$\mathrm{Sometimes}\:\mathrm{just}\:\mathrm{use}\:\mathrm{your}\:\mathrm{brain}\:\&\:\mathrm{experience} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{g}\left({x}\right)}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}\right]=\frac{{g}'\left({x}\right)\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}{g}\left({x}\right)}{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${g}'\left({x}\right)\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}{g}\left({x}\right)=\mathrm{1}\:\Rightarrow\:{g}\left({x}\right)={x} \\ $$$$\int\frac{{dx}}{\:\sqrt{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }}=\frac{{x}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}+{C} \\ $$
Answered by Frix last updated on 03/Nov/24
![∫(dx/((4x^2 +1)^(3/2) )) =^([t=tan^(−1) 2x]) (1/2)∫cos t dt=(1/2)sin t = =(x/( (√(4x^2 +1))))+C](https://www.tinkutara.com/question/Q213379.png)
$$\int\frac{{dx}}{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\overset{\left[{t}=\mathrm{tan}^{−\mathrm{1}} \:\mathrm{2}{x}\right]} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{cos}\:{t}\:{dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{t}\:= \\ $$$$=\frac{{x}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}+{C} \\ $$
Answered by Frix last updated on 03/Nov/24
![∫(dx/((4x^2 +1)^(3/2) )) =^([t=2x+(√(4x^2 +1))]) ∫((2t)/((t^2 +1)^2 ))dt= =−(1/(t^2 +1))=(x/( (√(4x^2 +1))))+C](https://www.tinkutara.com/question/Q213378.png)
$$\int\frac{{dx}}{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\overset{\left[{t}=\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\right]} {=}\:\int\frac{\mathrm{2}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{x}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}+{C} \\ $$