Advertisements
Advertisements
प्रश्न
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
बेरीज
Advertisements
उत्तर
\[\text{ Let I }= \int\frac{x^3}{\sqrt{x^8 + 2^2}}dx\]
\[ = \int\frac{x^3}{\sqrt{\left( x^4 \right)^2 + 2^2}}dx\]
\[\text{ Putting x}^4 = t\]
\[ \Rightarrow 4 x^3 \text{ dx }= dt\]
\[ \Rightarrow x^3 \cdot dx = \frac{dt}{4}\]
\[ \therefore I = \frac{1}{4}\int\frac{1}{\sqrt{t^2 + 2^2}}dt\]
\[ = \frac{1}{4} \text{ ln} \left| t + \sqrt{t^2 + 4} \right| + C\]
\[ = \frac{1}{4} \text{ ln }\left| x^4 + \sqrt{x^8 + 4} \right| + C ...........\left[ \because t = x^4 \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
\[\int \cos^{- 1} \left( \sin x \right) dx\]
`∫ cos ^4 2x dx `
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
\[\int\frac{1}{x (3 + \log x)} dx\]
\[\int\frac{x + 1}{x \left( x + \log x \right)} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]
\[\ ∫ x \text{ e}^{x^2} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int\frac{dx}{e^x + e^{- x}}\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int x^2 \sin^2 x\ dx\]
\[\int\frac{x^2 \tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int \sin^3 \sqrt{x}\ dx\]
\[\int x \cos^3 x\ dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{x + 1}{x^2 + 4x + 5} \text{ dx}\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{4 x^2 + 5x + 6} \text{ dx}\]
\[\int\frac{1 + x^2}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
