Mathematics

Value of $$\int \:e^x\:\frac{\left(x^2+3x+3\right)}{\left(x+2\right)^2}dx\:$$ is 


ANSWER

$$e^x\left[\dfrac{x+1}{x+2}\right]+c$$


SOLUTION
$$\int{{e}^{x}\left(\dfrac{{x}^{2}+3x+3}{{\left(x+2\right)}^{2}}\right)dx}$$
$$=\int{{e}^{x}\left(1-\dfrac{x+1}{{\left(x+2\right)}^{2}}\right)dx}$$
$$=\int{{e}^{x}dx}-\int{{e}^{x}\left(\dfrac{x+1}{{\left(x+2\right)}^{2}}\right)dx}$$
$$=\int{{e}^{x}dx}-\int{{e}^{x}\left(\dfrac{x+2-1}{{\left(x+2\right)}^{2}}\right)dx}$$ take $$1=2-1$$
$$=\int{{e}^{x}dx}-\int{{e}^{x}\left(\dfrac{x+2}{{\left(x+2\right)}^{2}}-\dfrac{1}{{\left(x+2\right)}^{2}}\right)dx}$$
$$=\int{{e}^{x}dx}-\int{{e}^{x}\left(\dfrac{1}{x+2}-\dfrac{1}{{\left(x+2\right)}^{2}}\right)dx}$$
Let $$f\left(x\right)=\dfrac{1}{x+2}$$ then $${f}^{\prime}{\left(x\right)}=\dfrac{-1}{{\left(x+2\right)}^{2}}$$
We have $$\int{{e}^{x}\left(f\left(x\right)+{f}^{\prime}\left(x\right)\right)}={e}^{x}f\left(x\right)+c$$
Using this we have $$\int{{e}^{x}\left(\dfrac{1}{x+2}-\dfrac{1}{{\left(x+2\right)}^{2}}\right)dx}={e}^{x}\left(\dfrac{1}{x+2}\right)+c$$
$$\therefore \int{{e}^{x}\left(\dfrac{{x}^{2}+3x+3}{{\left(x+2\right)}^{2}}\right)dx}$$
$$=\int{{e}^{x}dx}-\int{{e}^{x}\left(\dfrac{1}{x+2}-\dfrac{1}{{\left(x+2\right)}^{2}}\right)dx}$$
$$={e}^{x}-{e}^{x}\left(\dfrac{1}{x+2}\right)+c$$
$$={e}^{x}\left(1-\left(\dfrac{1}{x+2}\right)\right)+c$$
$$={e}^{x}\left(\dfrac{x+2-1}{x+2}\right)+c$$
$$={e}^{x}\left(\dfrac{x+1}{x+2}\right)+c$$
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Single Correct Hard Published on 17th 09, 2020
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