Advertisements
Advertisements
प्रश्न
\[\int\sqrt{1 + e^x} . e^x dx\]
बेरीज
उत्तर
\[\int\sqrt{1 + e^x} \cdot e^x dx\]
\[\text{Let 1 }+ e^x = t\]
\[ \Rightarrow e^x = \frac{dt}{dx}\]
\[ \Rightarrow e^x dx = dt\]
\[Now, \int\sqrt{1 + e^x} \cdot e^x dx\]
\[ = \int\sqrt{t} \cdot dt\]
\[ = \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} + C\]
\[ = \frac{2}{3} t^\frac{3}{2} + C\]
\[ = \frac{2}{3} \left( 1 + e^x \right)^\frac{3}{2} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\frac{1}{x (3 + \log x)} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \sec^4 2x \text{ dx }\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]
\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]
\[\int x^3 \cos x^2 dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{1}{\sec x + cosec x}\text{ dx }\]
\[\int \log_{10} x\ dx\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
