Advertisements
Advertisements
प्रश्न
\[\int\frac{x^2}{x^6 + a^6} dx\]
योग
उत्तर
\[\int\frac{x^2}{x^6 + a^6}dx\]
\[ \Rightarrow \int\frac{x^2 dx}{\left( x^3 \right)^2 + \left( a^3 \right)^2}\]
\[\text{ let } x^3 = t\]
\[ \Rightarrow 3 x^2 dx = dt\]
\[ \Rightarrow x^2 dx = \frac{dt}{3}\]
\[Now, \int\frac{x^2}{x^6 + a^6}dx\]
\[ = \frac{1}{3}\int\frac{dt}{t^2 + \left( a^3 \right)^2}\]
\[ = \frac{1}{3 a^3} \tan^{- 1} \left( \frac{t}{a^3} \right) + C\]
\[ = \frac{1}{3 a^3} \tan-^1 \left( \frac{x^3}{a^3} \right) + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{1}{1 - \sin x} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{x}{x^2 + 3x + 2} dx\]
\[\int\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1}\text{ dx }\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int x \sin^3 x\ dx\]
\[\int x \cos^3 x\ dx\]
\[\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \frac{x - 1}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int e^x \frac{1 + x}{\left( 2 + x \right)^2} \text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int\frac{\log \left( 1 - x \right)}{x^2} \text{ dx}\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\frac{\sin x + \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]
