Advertisements
Advertisements
प्रश्न
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
योग
उत्तर
` ∫ {x dx}/{\sqrt{x^4 + a^4}} `
` ∫ {x dx}/\sqrt{(x^2)^2 + (a^2)^2}`
` \text{ let} x^2 = t `
\[ \Rightarrow\text{ 2x dx } = dt\]
\[ \Rightarrow\text{ x dx } = \frac{dt}{2}\]
Now, ` ∫ {x dx}/\sqrt{(x^2)^2 + (a^2)^2}`
\[ = \frac{1}{2}\int\frac{dt}{\sqrt{t^2 + \left( a^2 \right)^2}}\]
\[ = \frac{1}{2} \text{ log }\left| t + \sqrt{t^2 + a^4} \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| x^2 + \sqrt{x^4 + a^4} \right| + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int\frac{2x + 3}{\left( x - 1 \right)^2} dx\]
\[\int \cos^2 \frac{x}{2} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]
\[\int\sqrt{1 + e^x} . e^x dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{1}{\sqrt{a^2 - b^2 x^2}} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{x + 1}{x^2 + x + 3} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]
\[\int x \cos x\ dx\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int x \cos^2 x\ dx\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]
\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]