Mathematics

# Solve $\displaystyle\int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}} \,dx\,$

##### SOLUTION
$\begin{array}{l}\displaystyle\int {\dfrac{{dx}}{{\sqrt {8 + 3x - {x^2}} }}} \\ = \displaystyle\int {\dfrac{{dx}}{{ - \sqrt {{x^2} - 3x - 8} }}} \\ = \displaystyle\int {\dfrac{{dx}}{{ - \sqrt {\left( {{x^2} - 2.\dfrac{3}{2}.x + \dfrac{9}{4}} \right) - \dfrac{9}{4} - 8} }}} \\ = \displaystyle\int {\dfrac{{dx}}{{ - \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \dfrac{{41}}{4}} }}} \\ = \displaystyle\int {\dfrac{{dx}}{{\sqrt {\dfrac{{41}}{4} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} }}} \\ = \displaystyle\int {\dfrac{{dx}}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} }}} \\ = {\sin ^{ - 1}}\dfrac{{x - \dfrac{3}{2}}}{{\dfrac{{\sqrt {41} }}{2}}} + C\\ = {\sin ^{ - 1}}\left( {\dfrac{{2x - 3}}{{\sqrt {41} }}} \right) + C\end{array}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
If $f\left (\dfrac {x - 4}{x + 2}\right ) = 2x + 1, (x\epsilon R - \left \{1, -2\right \})$m then $\int f(x) dx$ is equal to
(where $C$ is a constant of integration).
• A. $12\log_{e}|1 - x| + 3x + C$
• B. $-12\log_{e}|1 - x| - 3x + C$
• C. $-12\log_{e}|1 - x| + 3x + C$
• D. $12\log_{e}|1 - x| - 3x + C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of the integral $\displaystyle \int_{0}^{\dfrac{\pi ^{2}}{4}}\sin \sqrt{x}\: dx$ is
• A. $1$
• B. $1/2$
• C. $3/2$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $f\left ( x \right )$ is a monotonic and differentiable function on $\left [ a,b \right ]$,
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1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
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$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$