Advertisements
Advertisements
प्रश्न
\[\ ∫ x \text{ e}^{x^2} dx\]
योग
उत्तर
\[\int x . e^{x^2} dx\]
\[\text{Let x}^2 = t\]
\[ \Rightarrow \text{2x dx} = dt\]
\[ \Rightarrow \text{x dx} = \frac{dt}{2}\]
\[Now, \int x . e^{x^2} dx\]
\[ = \frac{1}{2}\int e^t dt\]
\[ = \frac{1}{2} e^t + C\]
\[ = \frac{1}{2} e^{x^2} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int \cos^2 \text{nx dx}\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int\frac{1}{\left( x + 1 \right)\left( x^2 + 2x + 2 \right)} dx\]
\[\int\frac{x^2}{\sqrt{x - 1}} dx\]
\[\int \sec^4 2x \text{ dx }\]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]
` ∫ sin x log (\text{ cos x ) } dx `
` ∫ x tan ^2 x dx
\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int \sin^4 2x\ dx\]
\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\]
\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]
\[\int \cot^5 x\ dx\]
\[\int \cos^5 x\ dx\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int\frac{\log x}{x^3} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
