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Question Number 39135 by maxmathsup by imad last updated on 02/Jul/18

c a l c u l a t e A (λ) = \int_{0}^{λ} \frac{l n (x + \sqrt{1 + x^{2})}}{\sqrt{1 + x^{2}}} d x

2) c a l c u l a t e \int_{0}^{1} \frac{l n (x + \sqrt{1 + x^{2})}}{\sqrt{1 + x^{2}}} d x

Commented by math khazana by abdo last updated on 03/Jul/18

l e t i n t e g r a t e b y p a r t s u^{^{'}} = \frac{1}{\sqrt{1 + x^{2}}} a n d v = l n (x + \sqrt{1 + x^{2})}

A (λ) = {[{l n}^{2} (x + \sqrt{1 + x^{2}})]}_{0}^{λ} - \int_{0}^{λ} \frac{l n (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} d x

\Rightarrow 2 A (λ) = {l n}^{2} (λ + \sqrt{1 + λ^{2}}) \Rightarrow

A (λ) = \frac{1}{2} {l n}^{2} (λ + \sqrt{1 + λ^{2}})

2) \int_{0}^{1} \frac{l n (1 + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}) d x = A (1) = \frac{1}{2} {l n}^{2} (2 + \sqrt{2}) .

Commented by math khazana by abdo last updated on 03/Jul/18

A (1) = \frac{1}{2} {l n}^{2} (1 + \sqrt{2})

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

1) t = l n (x + \sqrt{1 + x^{2}})

\frac{d t}{d x} = \frac{1 + \frac{2 x}{2 \sqrt{1 + x^{2}}}}{x + \sqrt{1 + x^{2}}} = \frac{1}{\sqrt{1 + x^{2}}}

\int_{0}^{l n (λ + \sqrt{1 + λ^{2}})} t d t

∣ \frac{t^{2}}{2} ∣_{0}^{l n (λ + \sqrt{1 + λ^{2})}}

\frac{1}{2} {l n {(λ + \sqrt{1 + λ^{2})}}}^{2}

2) \frac{1}{2} {l n ∣ 1 + \sqrt{2} ∣}^{2}