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Question: The value of \(\int_{–2}^{2}\left\{ p\mathcal{l}n\left( \frac{1 + x}{1–x} \right) + q\mathcal{l}n\le...

The value of ∫–22{pln(1+x1–x)+qln(1–x1+x)–2+r}\int_{–2}^{2}\left\{ p\mathcal{l}n\left( \frac{1 + x}{1–x} \right) + q\mathcal{l}n\left( \frac{1–x}{1 + x} \right)^{–2} + r \right\}dx

depends on -

A

p

B

q

C

r

D

q & p

Answer

r

Explanation

Solution

∫–22{pln(1+x1–x)+qln(1–x1+x)–2+γ}\int_{–2}^{2}\left\{ p\mathcal{l}n\left( \frac{1 + x}{1–x} \right) + q\mathcal{l}n\left( \frac{1–x}{1 + x} \right)^{–2} + \gamma \right\}dx

= p ∫–22ln(1+x1–x)\int_{–2}^{2}{\mathcal{l}n\left( \frac{1 + x}{1–x} \right)}dx–2q∫–22ln(1–x1+x)\int_{–2}^{2}{\mathcal{l}n\left( \frac{1–x}{1 + x} \right)}dx + ∫–22γdx\int_{–2}^{2}{\gamma dx}

= p × 0 – 2q × 0 + 4g = 4 g