# Evaluate the definite integral. \int_{0}^{2}\frac{x^{2}-2}{x+1}dx

Evaluate the definite integral.
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}{\left.{d}{x}\right.}$$

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Julie Mathew
Step 1
Using the following integrations to solve:
$$\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C}$$
$$\displaystyle\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$$
where C is the constant when indefinite integral is there, but in question we have definite integral so after solving the integration, given limits have to be substituted.
Step 2
Solving the integration as,
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{2}}}}{\left({\left({x}-{1}\right)}-{\left({\frac{{{1}}}{{{x}+{1}}}}\right)}\right)}{\left.{d}{x}\right.}$$
Now integrating it using the formulas,
$$\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}{\left.{d}{x}\right.}={{\left[{\frac{{{x}^{{{2}}}}}{{{2}}}}-{x}-{\ln}{\left|{x}+{1}\right|}\right]}_{{{0}}}^{{{2}}}}={\frac{{{1}}}{{{2}}}}{\left({4}\right)}-{2}-{\ln}{\left|{3}\right|}$$
$$\displaystyle={2}-{2}-{\ln}{\left|{3}\right|}$$
$$\displaystyle=-{\ln}{\left|{3}\right|}$$
###### Not exactly what you’re looking for?
Troy Lesure
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
Step 2: In this case, $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}-{2}}}{{{x}+{1}}}}$$. Find its integral.
$$\displaystyle{\frac{{{x}^{{{2}}}}}{{{2}}}}-{x}-{\ln{{\left({x}+{1}\right)}}}{{\mid}_{{{0}}}^{{{2}}}}$$
Step 3: Since $$\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into F(2)−F(0):
$$\displaystyle{\left({\frac{{{2}^{{{2}}}}}{{{2}}}}-{2}-{\ln{{\left({2}+{1}\right)}}}\right)}-{\left({\frac{{{0}^{{{2}}}}}{{{2}}}}-{0}-{\ln{{\left({0}+{1}\right)}}}\right)}$$
Step 4: Simplify.
$$\displaystyle-{\ln{{3}}}$$