∫ 3x 1 ( x+1 ) #160; ^ 2 dx let, 3x 1 ( x+1 ) #160; ^ 2 = A (x+1) + B ( x+1 ) #160; ^ 2 ⇒ 3x 1=A( x+1 ) +B ⇒ 3x 1=Ax+(A+B) compa ...

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Property 3

Solve : ∫3x...

Question

Solve : $\int \frac{3 x - 1}{(x + 1)^{2}} d x = ?$

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Solution

$\int \frac{3 x - 1}{{(x + 1)}^{2}} d x l e t, \frac{3 x - 1}{{(x + 1)}^{2}} = \frac{A}{(x + 1)} + \frac{B}{{(x + 1)}^{2}} \Rightarrow 3 x - 1 = A (x + 1) + B \Rightarrow 3 x - 1 = A x + (A + B)$
comparing both side we get,
$A = 3 a n d A + B = - 1 \Rightarrow B = - 4 ∴ \int \frac{3 x - 1}{{(x + 1)}^{2}} d x = \int [\frac{3}{(x + 1)} - \frac{4}{{(x + 1)}^{2}}] d x = 3 log (x + 1) + \frac{4}{(x + 1)} + C = log {(x + 1)}^{2} + \frac{4}{(x + 1)} + C$

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p

q

\frac{6 q - p}{6}

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