∫_1 ^( ∞) ((tan^(−1) (x))/x^2 ) dx ?


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Question Number 122976 by bemath last updated on 21/Nov/20

$\int_{1}^{\infty} \frac{\tan^{- 1} (x)}{x^{2}} d x ?$
	Answered by mnjuly1970 last updated on 21/Nov/20

	$s o l u t i o n ::$ $\frac{1}{x} = t \Rightarrow \frac{- d x}{x^{2}} = d t$ $\int_{0}^{1} {t a n}^{- 1} (\frac{1}{t}) d t = \frac{π}{2} - \int_{0}^{1} {t a n}^{- 1} (t) d t$ $\frac{π}{2} - {[t ({t a n}^{- 1} (t))]}_{0}^{1} + \int_{0}^{1} \frac{t}{1 + t^{2}} d t$ $= \frac{π}{2} - \frac{π}{4} + \frac{1}{2} l n (2) = \frac{π}{4} + \frac{1}{2} l n (2) ✓$
	Answered by TANMAY PANACEA last updated on 21/Nov/20

	$x = t a n a \to d x = {s e c}^{2} a d a$ $\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{a \times {s e c}^{2} a}{{t a n}^{2} a}$ $a \int \frac{d (t a n a)}{{t a n}^{2} a} - \int [\frac{d a}{d a} \int \frac{d (t a n a)}{{t a n}^{2} a}] d a$ $a \times \frac{- 1}{t a n a} + \int \frac{d a}{t a n a}$ $\frac{- a}{t a n a} + l n (s i n a) + C$ $◼ ∣ \frac{- a}{t a n a} + l n s i n a {\underset{\frac{π}{4}}{∣}}^{\frac{π}{2}}$ $- \frac{\frac{π}{2}}{t a n \frac{π}{2}} + \frac{\frac{π}{4}}{t a n \frac{π}{4}} + l n s i n \frac{π}{2} - l n s i n \frac{π}{4}$ $= \frac{- \frac{π}{2}}{\infty} + \frac{\frac{π}{4}}{1} + l n 1 - l n (\frac{1}{\sqrt{2}})$ $\frac{π}{4} - l n (\frac{1}{\sqrt{2}}) = \frac{π}{4} - (l n 1 - l n \sqrt{2}) = \frac{π}{4} + l n \sqrt{2}$
	Answered by benjo_mathlover last updated on 21/Nov/20
	$∫_1 ^∞ ((tan^(−1) (x))/x^2 ) dx ? by parts { ((u=tan^(−1) (x)⇒du=(dx/(1+x^2 )))),((dv=x^(−2) dx ⇒v=−x^(−1) )) :} ν = [ −x tan^(−1) (x) ]_1 ^∞ +∫_1 ^∞ (dx/(x(1+x^2 ))) ν= 0−(−(π/4))+∫_0 ^∞ ((1/x)−(x/((1+x^2 ))))dx ν = (π/4)+[ ln (x)−(1/2)ln (1+x^2 )]_0 ^∞ ν = (π/4) +(1/2)[ln ((x^2 /(1+x^2 )))]_1 ^∞ ν = (π/4)+(1/2)(0−ln ((1/2))) ν = (π/4) + ln (√2) .$
	$\int_{1}^{\infty} \frac{\tan^{- 1} (x)}{x^{2}} d x ?$ $b y p a r t s {\begin{cases} u = \tan^{- 1} (x) \Rightarrow d u = \frac{d x}{1 + x^{2}} \\ d v = x^{- 2} d x \Rightarrow v = - x^{- 1} \end{cases}$ $ν = {[- x \tan^{- 1} (x)]}_{1}^{\infty} + \underset{1}{\int^{\infty}} \frac{d x}{x (1 + x^{2})}$ $ν = 0 - (- \frac{π}{4}) + \underset{0}{\int^{\infty}} (\frac{1}{x} - \frac{x}{(1 + x^{2})}) d x$ $ν = \frac{π}{4} + {[\ln (x) - \frac{1}{2} \ln (1 + x^{2})]}_{0}^{\infty}$ $ν = \frac{π}{4} + \frac{1}{2} {[\ln (\frac{x^{2}}{1 + x^{2}})]}_{1}^{\infty}$ $ν = \frac{π}{4} + \frac{1}{2} (0 - \ln (\frac{1}{2}))$ $ν = \frac{π}{4} + \ln \sqrt{2} .$
	Answered by mathmax by abdo last updated on 21/Nov/20

	$A = \int_{1}^{\infty} \frac{arctanx}{x^{2}} dx by parts we get$ $A = {[- \frac{arctanx}{x}]}_{1}^{\infty} + \int_{1}^{\infty} \frac{1}{x} \frac{dx}{1 + x^{2}} = \frac{π}{4} + \int_{1}^{\infty} (\frac{1}{x} - \frac{x}{1 + x^{2}}) dx$ $= \frac{π}{4} + {[\ln ∣ x ∣ - \ln (\sqrt{1 + x^{2}})]}_{1}^{\infty} = \frac{π}{4} + {[\ln ∣ \frac{x}{\sqrt{1 + x^{2}}} ∣]}_{1}^{\infty} = \frac{π}{4} - \ln (\frac{1}{\sqrt{2}})$ $= \frac{π}{4} + \ln (\sqrt{2}) = \frac{π}{4} + \frac{\ln (2)}{2}$
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