Advertisements
Advertisements
Question
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
Solution
I = \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\]
\[I = \int\frac{dx}{\sqrt{- \left( x^2 + 2x - 3 \right)}}\]
\[ = \int\frac{dx}{\sqrt{- \left( x^2 + 2x - 4 + 1 \right)}}\]
\[ = \int\frac{dx}{\sqrt{- \left[ \left( x^2 + 2x + 1 \right) - 2^2 \right]}}\]
\[= \int\frac{dx}{\sqrt{- \left[ \left( x + 1 \right)^2 - 2^2 \right]}}\]
\[ = \int\frac{dx}{\sqrt{2^2 - \left( x + 1 \right)^2}}\]
\[ = \sin^{- 1} \left( \frac{x + 1}{2} \right) + C\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} dx\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int x^3 \sin x^4 dx\]
\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{a^2 - b^2 x^2} dx\]
\[\int\frac{x^4 + 1}{x^2 + 1} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int\frac{1}{e^x + e^{- x}} dx\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
