Advertisements
Advertisements
Question
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Advertisements
Solution
\[\text{ Let I} = \int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
\[\text{ Let } \tan^{- 1} x = t\]
\[ \Rightarrow \frac{dx}{1 + x^2} = dt\]
\[ \therefore I = \int e^t dt\]
\[ = e^t + C\]
\[ = e^{{tan}^{- 1}} x + C\]
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_0^3dx/(9+x^2)`
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
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:
Write a value of
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
