Mathematics

The value of $$\int { { e }^{ x } } .\dfrac { { x }^{ 2 }+1 }{ { \left( x+1 \right)  }^{ 2 } } dx$$ is 


ANSWER

$${ e }^{ x }\left( \dfrac { x-1 }{ x+1 } \right) +C$$


SOLUTION
$$I=\displaystyle\int e^x\dfrac{x^2+1}{(x+1)^2}dx$$
$$=\displaystyle\int e^x\dfrac{x^2+1+2x-2x}{(x+1)^2}dx$$
$$=\displaystyle\int e^x\dfrac{(x+1)^2-2x}{(x+1)^2}dx$$
$$=\displaystyle\int e^x\left\{1-\dfrac{2x}{(x+1)^2}\right\}dx$$
$$=\displaystyle\int e^xdx-\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$$.
$$=e^x-I_1+e$$
Let $$I_1=\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$$
Here $$\dfrac{2x}{(x+1)^2}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+1)^2}$$
$$2x=A(x+1)+B$$
If $$x=-1$$,
$$2(-1)=B$$
$$\Rightarrow B=-2$$
If $$x=1$$,
$$2=2A+(-2)$$
$$2A=2+2$$
$$A=\dfrac{4}{2}=2$$
$$\therefore \dfrac{2x}{(x+1)^2}=\dfrac{2}{x+1}-\dfrac{2}{(x+1)^2}$$
$$\therefore I_1=\displaystyle\int e^x\left\{\dfrac{2}{x+1}-\dfrac{2}{(x+1)^2}\right\}dx$$
$$=\displaystyle\int 2e^x\left\{\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\right\}dx$$
$$=2\displaystyle\int e^x\left\{\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\right\}dx$$
Which is of the form $$\displaystyle\int e^x[f(x)+f'(x)]dx$$
$$=2e^x\cdot \left(\dfrac{1}{x+1}\right)$$   $$[\because\displaystyle\int e^x[f(x)+f'(x)]dx=e^xf(x)+c]$$
$$=\dfrac{2e^x}{x+1}$$
$$\therefore I=e^x-\dfrac{2e^x}{x+1}+c$$
$$=e^x\left(1-\dfrac{2}{x+1}\right)+c$$
$$=e^x\left\{\dfrac{x+1-2}{x+1}\right\}+c$$
$$\therefore I=e^x\left(\dfrac{x-1}{x+1}\right)+c$$.
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