The correct option is C 2x+1 3(x+1)^2+2(x+1)^3 Let 2x^2+x+1(x+1)^3=Ax+1+B(x+1)^2+C(x+1)^3 Then making denominators equal on both sides 2x^2+x+1=A( ...

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Integration by Partial Fractions

2x2+x+1x+13=

Question

$\frac{2 x^{2} + x + 1}{(x + 1)^{3}} =$

\frac{2}{x + 1} + \frac{3}{(x + 1)^{2}} - \frac{2}{(x + 1)^{3}}

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\frac{2}{x + 1} - \frac{3}{(x + 1)^{2}} + \frac{2}{(x + 1)^{3}}

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\frac{2}{x + 1} + \frac{3}{(x + 1)^{2}} + \frac{2}{(x + 1)^{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{x + 1} - \frac{3}{(x + 1)^{2}} - \frac{2}{(x + 1)^{3}}

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Solution

The correct option is C $\frac{2}{x + 1} - \frac{3}{(x + 1)^{2}} + \frac{2}{(x + 1)^{3}}$
Let
$\frac{2 x^{2} + x + 1}{(x + 1)^{3}} = \frac{A}{x + 1} + \frac{B}{(x + 1)^{2}} + \frac{C}{(x + 1)^{3}}$
Then making denominators equal on both sides
$2 x^{2} + x + 1 = A (x + 1)^{2} + B (x + 1) + C$
Putting x=-1 we get C=2. Then
$A (x + 1)^{2} + B (x + 1) = 2 x^{2} + x - 1 = (x + 1) (2 x - 1)$
Dividing both sides by (x+1) and putting $x = - 1$ we get $B = - 3$
Now
$A (x + 1) + B = 2 x - 1$
we have $B = - 3$ and put x=0 above
$\Rightarrow A - 3 = 0 - 1$
$∴ A = 2$

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2x2+x+1(x+1)3=

$\frac{2 x^{2} + x + 1}{(x + 1)^{3}} =$