If (dx)/((x^2-2x+10)^2)=A^-1(x-13)+f(x)x^2-2x+10+C where C is a constant of integration, t…

Question

If $\displaystyle\int\frac{dx}{(x^2-2x+10)^2}=A	an^{-1}\left(\dfrac{x-1}3ight)+\dfrac{f(x)}{x^2-2x+10}+C$ where C is a constant of integration, then :

Accepted Answer

$$\int {{{dx} \over {{{({x^2} - 2x + 10)}^2}}} = \int {{{dx} \over {{{({{(x - 1)}^2} + 9)}^2}}}} } $$

$$Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)$$

$$ \Rightarrow \tan \theta  = {{x - 1} \over 3}$$

On Differentiating ...(i)

2(x -- 1)dx = 18tan$$\theta $$ sec2$$\theta $$ d$$\theta $$

$$ \therefore I = \int {{{18\tan \theta {{\sec }^2}\theta d\theta } \over {2 \times 3\tan \theta  \times 81{{\sec }^4}\theta }}} $$

$$I = {1 \over {27}}\int {{{\cos }^2}\theta \,d\theta }  = {1 \over {27}} \times {1 \over 2}\int {\left( {1 + \cos 2\theta } \right)} d\theta $$

$$I = {1 \over {54}}\left\{ {\theta  + {{\sin 2\theta } \over 2}} \right\} + c$$

$$I = {1 \over {54}}\left\{ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {1 \over 2} \times {{2\left( {{{x - 1} \over 3}} \right)} \over {1 + {{\left( {{{x - 1} \over 3}} \right)}^2}}}} \right\} + c$$

$$I = {1 \over {54}}\left[ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{3(x - 1)} \over {{x^2} - 2x + 10}}} \right] + c$$

then $$A = {1 \over {54}}$$

f(x) = 3(x -- 1)

Solutions · 1