Advertisements
Advertisements
प्रश्न
` ∫ { x^2 dx}/{x^6 - a^6} dx `
बेरीज
Advertisements
उत्तर
\[\int\frac{x^2 dx}{x^6 - a^6}\]
\[\text{ let } x^3 = t\]
\[ \Rightarrow 3 x^2 \text{ dx } = dt\]
\[ \Rightarrow x^2 dx = \frac{dt}{3}\]
\[Now, \int\frac{x^2 dx}{x^6 - a^6}\]
\[ = \frac{1}{3}\int\frac{dt}{t^2 - \left( a^3 \right)^2}\]
\[ = \frac{1}{3} \times \frac{1}{2 a^3} \text{ log } \left| \frac{t - a^3}{t + a^3} \right| + C\]
\[ = \frac{1}{6 a^3} \text{ log }\left| \frac{x^3 - a^3}{x^3 + a^3} \right| + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int \left( 3x + 4 \right)^2 dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int \sin^5\text{ x }\text{cos x dx}\]
\[\int\frac{\cos^5 x}{\sin x} dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
\[\int\frac{e^x}{e^{2x} + 5 e^x + 6} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{x^3}{x^4 + x^2 + 1}dx\]
\[\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int x e^x \text{ dx }\]
`int"x"^"n"."log" "x" "dx"`
\[\int\cos\sqrt{x}\ dx\]
` ∫ x tan ^2 x dx
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]
\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
\[\int\frac{x^4 + x^2 - 1}{x^2 + 1} \text{ dx}\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int \sec^4 x\ dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
