Advertisements
Advertisements
प्रश्न
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Advertisements
उत्तर
\[\text{ Let I} = \int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\text{ Let } \tan^{- 1} x = t\]
\[ \Rightarrow \frac{dx}{1 + x^2} = dt\]
\[ \therefore I = \int e^t dt\]
\[ = e^t + C\]
\[ = e^{{tan}^{- 1}} x + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^3dx/(9+x^2)`
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
