[ On divinding ( x^ 2 +1 ) by ( x^ 2 +2x+1 ) , we get; ( x^ 2 +1 ) #160; /( x+1 ) #160; ^ 2 ={ 1 2x /( x+1 ) # ...

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Integration by Partial Fractions

∫ x2 + 1/ x +...

Question

$\int \frac{(x^{2} + 1)}{{(x + 1)}^{2}} d x$

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Solution

$\begin{matrix} O n d i v i n d i n g (x^{2} + 1) b y (x^{2} + 2 x + 1), w e g e t \frac{(x^{2} + 1)}{{(x + 1)}^{2}} = {1 - \frac{2 x}{{(x + 1)}^{2}}} L e t \frac{2 x}{{(x + 1)}^{2}} = \frac{A}{(x + 1)} + \frac{B}{{(x + 1)}^{2}} 2 x = A (x + 1) + B . - - - - (1) O n e q u a t i n g t h e c o e f f i c i e n t s o f x, w e g e t A = 2 O n e q u a t i n g c o n s t a n t t e r m, w e g e t A + B = 0 \Rightarrow B = - A = - 2 ∴ \frac{2 x}{{(x + 1)}^{2}} = \frac{2}{(x + 1)} - \frac{2}{{(x + 1)}^{2}} ∴ I = \int {1 - \frac{2 x}{{(x + 1)}^{2}}} = \int {1 - \frac{2}{(x + 1)} + \frac{2}{{(x + 1)}^{2}}} d x x - 2 log | x + 1 | - \frac{2}{(x + 1)} + C . \end{matrix}$

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

I_{2} = \int_{1}^{2} \frac{d x}{x}

I_{1}

I_{2}

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