Advertisements
Advertisements
Question
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Solution
\[ = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c\]
\[\text{ Hence,} \int\frac{1}{x^2 + 16}dx = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c\]
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_0^3dx/(9+x^2)`
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
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:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)