I-0-lnx-x-2-2x-4-dx-put-x-1-3-tan-I-0-pi-2-ln-3-tan-1-3-sec-2-3-sec-2-d-I-1-3-0-pi-2-ln-3-tan-1-d-I-1-3-0-pi-2-ln-3-sin-cos-

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}