Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 145231 by mathdanisur last updated on 03/Jul/21

$$1<a⩽bthenfind$$ $$\underset{\mathit{a}}{\stackrel{\mathit{b}}{\int }}{tan}^{-1}\left(\frac{3x}{1-2x^2}\right)dx=?$$

Commented byDwaipayan Shikari last updated on 03/Jul/21

$${tan}^{-1}\frac{3x}{1-2x^2}={tan}^{-1}x+{tan}^{-1}2x$$ $$\int {tan}^{-1}xdx={xtan}^{-1}x-\int \frac{x}{1+x^2}dx={xtan}^{-1}x-\frac{1}{2}log(1+x^2)$$ $$Nowputthese...$$

Commented bymathdanisur last updated on 03/Jul/21

$$thankyouSer,answer?$$

Commented byDwaipayan Shikari last updated on 03/Jul/21

$$b({tan}^{-1}b+{tan}^{-1}2b)-a({tan}^{-1}a+{tan}^{-1}2a)-\frac{1}{2}log(\frac{1+b^2}{1+a^2}.\sqrt{\frac{1+4a^2}{1+4b^2}})$$

Answered by mathmax by abdo last updated on 04/Jul/21

$$\mathrm{\Psi }={\int }_{\mathrm{a}}^{\mathrm{b}}\arctan \left(\frac{3\mathrm{x}}{1-{2\mathrm{x}}^2}\right)\mathrm{dx}\mathrm{by}\mathrm{parts}$$ $$\mathrm{\Psi }={\left[\mathrm{xarctan}\left(\frac{3\mathrm{x}}{1-{2\mathrm{x}}^2}\right)\right]}_{\mathrm{a}}^{\mathrm{b}}-{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}\times \frac{{\left(\frac{3\mathrm{x}}{1-{2\mathrm{x}}^2}\right)}^{{}^{\prime }}}{1+{\left(\frac{3\mathrm{x}}{1-{2\mathrm{x}}^2}\right)}^2}\mathrm{dx}$$ $$\mathrm{but}{(....)}^{{}^{\prime }}=\frac{3(1-{2\mathrm{x}}^2)-3\mathrm{x}(-4\mathrm{x})}{{(1-{2\mathrm{x}}^2)}^2}=\frac{3-{6\mathrm{x}}^2+{12\mathrm{x}}^2}{{(1-{2\mathrm{x}}^2)}^2}=\frac{3+{6\mathrm{x}}^2}{{(1-{2\mathrm{x}}^2)}^2}\Rightarrow$$ $$\mathrm{\Psi }=\mathrm{barctan}\left(\frac{3\mathrm{b}}{1-{2\mathrm{b}}^2}\right)-\mathrm{aarctan}\left(\frac{3\mathrm{a}}{1-{2\mathrm{a}}^2}\right)$$ $$-{\int }_{\mathrm{a}}^{\mathrm{b}}\frac{3\mathrm{x}+{6\mathrm{x}}^3}{{(1-{2\mathrm{x}}^2)}^2(1+\frac{{9\mathrm{x}}^2}{{(1-{2\mathrm{x}}^2)}^2})}\mathrm{dx}$$ $$=.....-{\int }_{\mathrm{a}}^{\mathrm{b}}\frac{3\mathrm{x}+{6\mathrm{x}}^3}{{(1-{2\mathrm{x}}^2)}^2+{9\mathrm{x}}^2}\mathrm{dx}\mathrm{we}\mathrm{have}$$ $${\int }_{\mathrm{a}}^{\mathrm{b}}\frac{{6\mathrm{x}}^3+3\mathrm{x}}{{4\mathrm{x}}^4-{4\mathrm{x}}^2+1+{9\mathrm{x}}^2}\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{b}}\frac{{6\mathrm{x}}^3+3\mathrm{x}}{{4\mathrm{x}}^4+{5\mathrm{x}}^2+1}\mathrm{dx}$$ $${4\mathrm{x}}^4+{5\mathrm{x}}^2+1=0\Rightarrow {4\mathrm{u}}^2+5\mathrm{u}+1=0(\mathrm{u}={\mathrm{x}}^2)$$ $$\to \mathrm{\Delta }=25-16=9\Rightarrow {\mathrm{u}}_1=\frac{-5+3}{8}=-\frac{1}{4}\mathrm{and}{\mathrm{u}}_2=\frac{-5-3}{8}=-1\Rightarrow$$ $${4\mathrm{u}}^2+5\mathrm{u}+1=4(\mathrm{u}+\frac{1}{4})(\mathrm{u}+1)=(\mathrm{u}+1)(4\mathrm{u}+1)=({\mathrm{x}}^2+1)({4\mathrm{x}}^2+1)$$ $$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{{6\mathrm{x}}^3+3\mathrm{x}}{({\mathrm{x}}^2+1)({4\mathrm{x}}^2+1)}\Rightarrow$$ $$\mathrm{F}\left(\mathrm{x}\right)=\frac{\alpha \mathrm{x}+\beta }{{\mathrm{x}}^2+1}+\frac{\mathrm{cx}+\mathrm{d}}{{4\mathrm{x}}^2+1}\mathrm{determinatin}\mathrm{of}\mathrm{coefficient}\mathrm{is}\mathrm{eazy}\Rightarrow$$ $$\int \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\alpha }{2}\log ({\mathrm{x}}^2+1)+\beta \mathrm{arctanx}+\frac{\mathrm{c}}{8}\log ({4\mathrm{x}}^2+1)+\mathrm{d}\int \frac{\mathrm{dx}}{{4\mathrm{x}}^2+1}$$ $$\int \frac{\mathrm{dx}}{{4\mathrm{x}}^2+1}{=}_{2\mathrm{x}=\mathrm{t}}\frac{1}{2}\int \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}=\frac{1}{2}\arctan \left(2\mathrm{x}\right)+\mathrm{c}....$$ $$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com