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#integration____________​

Answer»

We're asked to EVALUATE,

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(x^<klux>2</klux>-4)\sqrt{x+1}}\ dx=\,?}

Let,

\displaystyle\longrightarrow\sf{u=\sqrt{x+1}}

\displaystyle\longrightarrow\sf{x=u^2-1}

\displaystyle\longrightarrow\sf{dx=2u\ du}

Also,

\displaystyle\longrightarrow\sf{x^2-4=(u^2-1)^2-4}

\displaystyle\longrightarrow\sf{x^2-4=u^4-2u^2-3}

\displaystyle\longrightarrow\sf{x^2-4=u^4+u^2-3u^2-3}

\displaystyle\longrightarrow\sf{x^2-4=u^2(u^2+1)-3(u^2+1)}

\displaystyle\longrightarrow\sf{x^2-4=(u^2+1)(u^2-3)}

Thus,

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(x^2-4)\sqrt{x+1}}\ dx=\int\dfrac{1}{(u^2+1)(u^2-3)u}\cdot2u\ du}

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(x^2-4)\sqrt{x+1}}\ dx=2\int\dfrac{1}{(u^2+1)(u^2-3)}\ du\quad\quad\dots(1)}

Let,

\displaystyle\longrightarrow\sf{\dfrac{1}{(u^2+1)(u^2-3)}=\dfrac{2Au+B}{u^2+1}+\dfrac{2Cu+D}{u^2-3}\quad\quad\dots(2)}

for some constants \sf{A} and \sf{B.}

\displaystyle\longrightarrow\sf{\dfrac{1}{(u^2+1)(u^2-3)}=\dfrac{(2Au+B)(u^2-3)+(2Cu+D)(u^2+1)}{(u^2+1)(u^2-3)}}

\displaystyle\longrightarrow\sf{(2Au+B)(u^2-3)+(2Cu+D)(u^2+1)=1}

\displaystyle\longrightarrow\sf{2Au^3-6Au+Bu^2-3B+2Cu^3+2Cu+Du^2+D=1}

\displaystyle\longrightarrow\sf{2(A+C)u^3+(B+D)u^2-2(3A-C)u+(D-3B)=1}

Equating corresponding coefficients,

\displaystyle\longrightarrow\sf{A+C=0}

\displaystyle\longrightarrow\sf{B+D=0}

\displaystyle\longrightarrow\sf{3A-C=0}

\displaystyle\longrightarrow\sf{D-3B=1}

Solving these FOUR equations we get,

\displaystyle\longrightarrow\sf{A=0}

\displaystyle\longrightarrow\sf{B=-\dfrac{1}{4}}

\displaystyle\longrightarrow\sf{C=0}

\displaystyle\longrightarrow\sf{D=\dfrac{1}{4}}

Then (2) becomes,

\displaystyle\longrightarrow\sf{\dfrac{1}{(u^2+1)(u^2-3)}=\dfrac{-\frac{1}{4}}{u^2+1}+\dfrac{\frac{1}{4}}{u^2-3}}

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(u^2+1)(u^2-3)}\ du=\int\left(\dfrac{-\frac{1}{4}}{u^2+1}+\dfrac{\frac{1}{4}}{u^2-3}\right)\ du}

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(u^2+1)(u^2-3)}\ du}=\ \ &\sf{-\dfrac{1}{4}\int\dfrac{1}{u^2+1}\ du+\dfrac{1}{4}\int\dfrac{1}{u^2-3}\ du}

We know that,

  • \displaystyle\sf{\int\dfrac{1}{x^2+1}=\tan^{-1}x}

  • \displaystyle\sf{\int\dfrac{1}{x^2-a^2}\ dx=\dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right|} for some constant \sf{a.}

So,

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(u^2+1)(u^2-3)}\ du=-\dfrac{1}{4}\tan^{-1}u+\dfrac{1}{8\sqrt3}\ln\left|\dfrac{u-\sqrt3}{u+\sqrt3}\right|}

So (1) will be,

\displaystyle\longrightarrow\sf{\int\dfrac{1}{(x^2-4)\sqrt{x+1}}\ dx=\dfrac{1}{4\sqrt3}\ln\left|\dfrac{u-\sqrt3}{u+\sqrt3}\right|-\dfrac{1}{2}\tan^{-1}u+c}

UNDOING \sf{u=\sqrt{x+1},} we get,

\displaystyle\begin{aligned}\longrightarrow\ \ &\sf{\int\dfrac{1}{(x^2-4)\sqrt{x+1}}\ dx}\\\\=\ \ &\underline{\underline{\sf{\dfrac{1}{4\sqrt3}\ln\left|\dfrac{\sqrt{x+1}-\sqrt3}{\sqrt{x+1}+\sqrt3}\right|-\dfrac{1}{2}\tan^{-1}\sqrt{x+1}+c}}}\end{aligned}


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