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  • Evaluate the integrals in Exercises 1 to 8 using substitution. dx/ (x+4-x^2)

Evaluate the integrals in Exercises 1 to 8 using substitution.

    Q6.    \int_0^2\frac{dx}{x + 4 - x^2}

Answers (1)

best_answer

By adjusting, the denominator can also be written as  (\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2
Now,
\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}
let x-1/2 = t\Rightarrow dx=dt
when x= 0 then t =-1/2 and when x =2 then t = 3/2
 

\\\Rightarrow\int_{-1/2}^{3/2}\frac{dt}{(\frac{\sqrt{17}}{2})^2-t^2}\\ =\frac{1}{2.\frac{\sqrt{17}}{2}}\log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}/2+3/2}{\sqrt{17}/2-3/2}-\log\frac{\sqrt{17}/2-1/2}{\sqrt{17}/2+1/2}]\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}+3}{\sqrt{17}-3/}.\frac{\sqrt{17}+1}{\sqrt{17}+1}]
\\ =\frac{1}{\sqrt{17}}[\log (\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}})]\\ =\frac{1}{\sqrt{17}}[\log (\frac{5+\sqrt{17}}{5-\sqrt{17}})]
On rationalisation, we get

=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}

Posted by

manish

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