Question

#integration
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Answer 1

We're asked to evaluate,

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

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,

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

Answer 2

Step-by-step explanation:

Answer:

{ln(x²-12+8√2)} /4 - ( 11+5√2)  tan^-1{x/(√8 - 2)}

Step-by-step explanation:

first find the partial fractions

x²-4/x⁴+24x²+16 = A/x² + 12-8√2 + B/x² - 12+8√2

x² - 4 =  A(x² - 12+8√2) + B(x² + 12 - 8√2 )

@ x = 12-8√2  , 2B(12-8√2) = 8-8√2 , B = -(1+5√2)

@ x = 12+8√2  , 2A(12+8√2) = 8+8√2 , A = √2 - 1

so far

I = ∫x²-4/x⁴+24x²+16 dx = ∫  A/(x² - 12+8√2) .dx + ∫B/(x² + 12 - 8√2 ) dx

∫Adx/{x² - (12-8√2) } = ∫Adx/{x² - (12-2√32) }  =  ∫Adx/{x² - (√8-2)² }

= A/2(√8 -2) {ln(x²-12+8√2)} = {ln(x²-12+8√2)} /4

∫Bdx/(x² + 12 - 8√2 ) dx = B ∫dx/{x² +  (√8-2 )² } = B ∫dx/{x² +  (√8 - 2 )² }

B/(√8 - 2 ) tan^-1{x/(√8 - 2)} = - ( 11+5√2)  tan^-1{x/(√8 - 2)}

therefore

I = ∫x²-4/x⁴+24x²+16 dx = ∫  A/(x² - 12+8√2) .dx + ∫B/(x² + 12 - 8√2 ) dx

I = {ln(x²-12+8√2)} /4 - ( 11+5√2)  tan^-1{x/(√8 - 2)}