Advertisements
Advertisements
प्रश्न
\[\int\left( x^e + e^x + e^e \right) dx\]
बेरीज
Advertisements
उत्तर
\[\int\left( x^e + e^x + e^e \right)dx\]
\[ = \int x^e dx + \int e^x dx + e^e \int1dx\]
\[ = \frac{x^{e + 1}}{e + 1} + e^x + x \cdot e^e + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{\sin x}{\sqrt{4 \cos^2 x - 1}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx }\]
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
\[\int\frac{1}{1 - \cot x} dx\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int x \text{ sin 2x dx }\]
\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} dx\]
\[\int\frac{1}{x \left( x^4 + 1 \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
\[\int\frac{x + 1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
If `int(2x^(1/2))/(x^2) dx = k . 2^(1/x) + C`, then k is equal to ______.
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
\[\int \tan^3 x\ dx\]
\[\int \tan^4 x\ dx\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int \tan^3 x\ \sec^4 x\ dx\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
