I=∫_0 ^∞ ((lnx)/(x^2 +2x+4)) dx put x+1=(√3) tanθ I=∫_0 ^(π/2) ((ln((√3) tanθ−1))/(3(sec^2 θ))) (√3) sec^2 θdθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)tanθ−1) dθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)sinθ−cosθ)−lncosθ dθ I=(1/( (√3)))∫_0 ^(π/2) ln(sin(θ−(π/6))+ln2−lncosθ dθ A=∫_(−2) ^6 ∣g(x)∣ dx ⇒let g(p)=0⇒ f(0)=p=2 A=∫_(−2) ^2 −g(x) dx+∫_2 ^6 g(x) dx ⇒put x=f(t) A=∫_(−1) ^0 −tf′(t) dt+∫_0 ^1 t f(t) dt A=∫_(−1) ^0 −3t^3 −3t dt+∫_0 ^1 3t^3 +3t dt A=2(((3t^4 )/4)+((3t^2 )/2))_0 ^1 =(9/2) lnx=(1/x)⇒xlnx=1(let x=α) ∫_1 ^α (1/x)−lnx dx=∫_α ^a lnx−(1/x) dx ⇒lnx−xlnx+x]_1 ^α =xlnx−x−lnx]_α ^a ⇒lnα−1+α−1=alna−a−lna−1+α +lnα ⇒alna−a−lna=−1 ⇒alna−lna=a−1⇒a=e s(t)=(1/2)∣OA^→ ×OB^→ ∣ s(t)=(1/2)∣(2,2,1)×(t,1,t+1)∣ s(t)=(1/2)∣(2t+1),(−2−t),(2−2t)∣ f(x)=(1/4)∫_0 ^x (2t+1)^2 +(t+2)^2 +(2−2t)^2 dt f(x)=(1/4)∫_0 ^x 9t^2 +9 dt=((3x^3 +9x)/4) A=(1/4)∫_0 ^6 (3x^3 +9x)dx=((3x^4 +18x^2 )/(16))]


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Question Number 118934 by Riteshgoyal last updated on 20/Oct/20

$I = \int_{0}^{\infty} \frac{l n x}{x^{2} + 2 x + 4} d x$ $p u t x + 1 = \sqrt{3} t a n θ$ $I = \int_{0}^{π / 2} \frac{l n (\sqrt{3} t a n θ - 1)}{3 ({s e c}^{2} θ)} \sqrt{3} {s e c}^{2} θ d θ$ $I = \frac{1}{\sqrt{3}} \int_{0}^{π / 2} l n (\sqrt{3} t a n θ - 1) d θ$ $I = \frac{1}{\sqrt{3}} \int_{0}^{π / 2} l n (\sqrt{3} s i n θ - c o s θ) - l n c o s θ d θ$ $I = \frac{1}{\sqrt{3}} \int_{0}^{π / 2} l n (s i n (θ - \frac{π}{6}) + l n 2 - l n c o s θ d θ$ $A = \int_{- 2}^{6} ∣ g (x) ∣ d x$ $\Rightarrow l e t g (p) = 0 \Rightarrow f (0) = p = 2$ $A = \int_{- 2}^{2} - g (x) d x + \int_{2}^{6} g (x) d x$ $\Rightarrow p u t x = f (t)$ $A = \int_{- 1}^{0} - {t f}^{'} (t) d t + \int_{0}^{1} t f (t) d t$ $A = \int_{- 1}^{0} - 3 t^{3} - 3 t d t + \int_{0}^{1} 3 t^{3} + 3 t d t$ $A = 2 {(\frac{3 t^{4}}{4} + \frac{3 t^{2}}{2})}_{0}^{1} = \frac{9}{2}$ $l n x = \frac{1}{x} \Rightarrow x l n x = 1 (l e t x = α)$ $\int_{1}^{α} \frac{1}{x} - l n x d x = \int_{α}^{a} l n x - \frac{1}{x} d x$ ${\Rightarrow {l n x - x l n x + x]}_{1}^{α} = x l n x - x - l n x]}_{α}^{a}$ $\Rightarrow l n α - 1 + α - 1 = a l n a - a - l n a - 1 + α$ $+ l n α$ $\Rightarrow a l n a - a - l n a = - 1$ $\Rightarrow a l n a - l n a = a - 1 \Rightarrow a = e$ $s (t) = \frac{1}{2} ∣ O \vec{A} \times O \vec{B} ∣$ $s (t) = \frac{1}{2} ∣ (2, 2, 1) \times (t, 1, t + 1) ∣$ $s (t) = \frac{1}{2} ∣ (2 t + 1), (- 2 - t), (2 - 2 t) ∣$ $f (x) = \frac{1}{4} \int_{0}^{x} {(2 t + 1)}^{2} + {(t + 2)}^{2} + {(2 - 2 t)}^{2} d t$ $f (x) = \frac{1}{4} \int_{0}^{x} 9 t^{2} + 9 d t = \frac{3 x^{3} + 9 x}{4}$ ${A = \frac{1}{4} \int_{0}^{6} (3 x^{3} + 9 x) d x = \frac{3 x^{4} + 18 x^{2}}{16}]}_{0}^{6}$ $A = \frac{3 . 6^{3} (6 + 1)}{16} = \frac{567}{2}$
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