∫2t/t^2 1dt ⇒∫2t/(t+1)(t 1)dt = A/t+1+B/t 1 Now, 2t = A(t 1)+B(t+1) A+B = 2 ⇒ 2B = 2 = B = A = 1 A+B = 0 A = B = ∫1/t+1dt+1/t ...

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Logarithms

∫2tt2-1dt

Question

$\int \frac{2 t}{t^{2} - 1} d t$

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Solution

$\int \frac{2 t}{t^{2} - 1} d t$
$\Rightarrow \int \frac{2 t}{(t + 1) (t - 1)} d t = \frac{A}{t + 1} + \frac{B}{t - 1}$
Now,
$2 t = A (t - 1) + B (t + 1)$
$A + B = 2 \Rightarrow 2 B = 2 = B = A = 1$
$- A + B = 0$
$A = B$
$= \int \frac{1}{t + 1} d t + \frac{1}{t - 1} d t$
$\Rightarrow l o g (t + 1) + l o g (t - 1) + c$
$\Rightarrow l o g (t + 1) (t - 1) + c$
$\Rightarrow l o g (t^{2} - 1) + c$

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