Question

integration of 1/(2-3x-x^2)​

Answer 1

TO FIND :–

$\\\bf \implies\int \dfrac{1}{2 - 3x - {x}^{2}}.dx =?$

SOLUTION :–

• Let the integration –

$\\\bf \implies I = \int \dfrac{1}{2 - 3x - {x}^{2}}dx$

$\\\bf \implies I = - \int \dfrac{1}{{x}^{2} + 3x - 2}dx$

$\\\bf \implies I = - \int \dfrac{1}{{x}^{2} +2x +x - 2}dx$

$\\\bf \implies I = - \int \dfrac{1}{{x}^{2} + 3x - 2 + \frac{9}{4} - \frac{9}{4} }dx$

$\\\bf \implies I = - \int \dfrac{1}{{x}^{2} + 3x + \frac{9}{4} - 2 - \frac{9}{4} }dx$

$\\\bf \implies I = - \int \dfrac{1}{ \bigg({x + \dfrac{3}{2} \bigg)}^{2} - \dfrac{17}{4} }dx$

$\\\bf \implies I = \int \dfrac{1}{ \bigg(\dfrac{ \sqrt{17}}{2} \bigg)^{2} - \bigg({x + \dfrac{3}{2} \bigg)}^{2} }dx$

• Using identity –

$\\\bf \longrightarrow \int \dfrac{1}{(a)^{2} - ({x)}^{2} }dx = \dfrac{1}{2a} log \bigg |\dfrac{a + x}{a - x} \bigg| + c$

• So that –

$\\\bf \implies I = \dfrac{1}{2\bigg(\dfrac{ \sqrt{17}}{2} \bigg)} log \left|\dfrac{\bigg(\dfrac{ \sqrt{17}}{2} \bigg) + x + \dfrac{3}{2} }{\bigg(\dfrac{ \sqrt{17}}{2} \bigg) - x - \dfrac{3}{2}} \right| + c$

$\\\bf \implies I = \dfrac{1}{\sqrt{17}}log \left|\dfrac{\dfrac{ \sqrt{17} + 3}{2} + x}{\dfrac{ \sqrt{17} - 3}{2}- x }\right| + c$

$\\\bf \implies I = \dfrac{1}{\sqrt{17}}log \left|\dfrac{\dfrac{ \sqrt{17} +3 +2x}{2} } {\dfrac{ \sqrt{17} - 3 -2x}{2}}\right| + c$

$\\\implies {\boxed{ \bf I = \dfrac{1}{\sqrt{17}}log \left|\dfrac{\sqrt{17} + 3 + 2x} {\sqrt{17} -3-2x}\right| + c}}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\int \dfrac{1}{2-3x-x^2}dx}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int \dfrac{1}{2-3x-x^2}dx=d\frac{1}{\sqrt{17}}\left(\ln \left|\dfrac{2x+3}{\sqrt{17}}+1\right|-\ln \left|\dfrac{2x+3}{\sqrt{17}}-1\right|\right)+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Complete the square } 2-3 x-x^{2}:-\left(x+\dfrac{3}{2}\right)^{2}+\dfrac{17}{4}$

$=\int \dfrac{1}{-\left(x+\dfrac{3}{2}\right)^2+\dfrac{17}{4}}dx$

$\text { Apply u - substitution: } u=x+\dfrac{3}{2}$

$=\int \dfrac{4}{-4u^2+17}du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=4\cdot \int \dfrac{1}{-4u^2+17}du$

$\text { Apply Integral Substitution: } u=\dfrac{\sqrt{17}}{2} v$

$=4\cdot \int \dfrac{1}{2\sqrt{17}\left(-v^2+1\right)}dv$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=4\cdot \dfrac{1}{2\sqrt{17}}\cdot \int \dfrac{1}{-v^2+1}dv$

$\mathrm{Use\:the\:common\:integral}:\quad \int \dfrac{1}{-v^2+1}dv=\dfrac{\ln \left|v+1\right|}{2}-\dfrac{\ln \left|v-1\right|}{2}$

$=4\cdot \dfrac{1}{2\sqrt{17}}\left(\dfrac{\ln \left|v+1\right|}{2}-\dfrac{\ln \left|v-1\right|}{2}\right)$

$\mathrm{Substitute\:back}$

$=4\cdot \dfrac{1}{2\sqrt{17}}\left(\dfrac{\ln \left|\dfrac{2}{\sqrt{17}}\left(x+\dfrac{3}{2}\right)+1\right|}{2}-\dfrac{\ln \left|\dfrac{2}{\sqrt{17}}\left(x+\dfrac{3}{2}\right)-1\right|}{2}\right)$

$\sf{Simplify\:\:4\cdot \dfrac{1}{2\sqrt{17}}\left(\dfrac{\ln \left|\dfrac{2}{\sqrt{17}}\left(x+\dfrac{3}{2}\right)+1\right|}{2}-\dfrac{\ln \left|\dfrac{2}{\sqrt{17}}\left(x+\dfrac{3}{2}\right)-1\right|}{2}\right)}$

$=\dfrac{1}{\sqrt{17}}\left(\ln \left|\dfrac{2x+3}{\sqrt{17}}+1\right|-\ln \left|\dfrac{2x+3}{\sqrt{17}}-1\right|\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$\boxed{\sf{=\dfrac{1}{\sqrt{17}}\left(\ln \left|\dfrac{2x+3}{\sqrt{17}}+1\right|-\ln \left|\dfrac{2x+3}{\sqrt{17}}-1\right|\right)+C}}$