`int (x^4 + x - 4)/(x^2 + 2) dx` Find the indefinite integral.

Tuesday, April 20, 2010

$\displaystyle\int\frac{{{{x}}^{{4}}+{x}-{4}}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$ Find the indefinite integral.

In Substitution Rule, we follow $\displaystyle\int{f{{\left({g{{\left({x}\right)}}}\right)}}}{g{'}}{\left({x}\right)}{\left.{d}{x}\right.}=\int{f{{\left({u}\right)}}}{d}{u}$ where we let $\displaystyle{u}={g{{\left({x}\right)}}}$ .

Before we use this, we look for possible way to simplify the function using math operation or algebraic techniques.

For the problem: $\displaystyle\int\frac{{{{x}}^{{4}}+{x}-{4}}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$ ,we expand first using long division.

$\displaystyle\frac{{{{x}}^{{4}}+{x}-{4}}}{{{{x}}^{{2}}+{2}}}={{x}}^{{2}}-{2}+\frac{{x}}{{{{x}}^{{2}}+{2}}}$

Applying $\displaystyle\int{\left({f{{\left({x}\right)}}}\pm{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\pm\int{g{{\left({x}\right)}}}{\left.{d}{x}\right.}:$

We get $\displaystyle\int{{x}}^{{2}}{\left.{d}{x}\right.}-\int{2}{\left.{d}{x}\right.}+\int\frac{{x}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}.$

$\displaystyle\int{{x}}^{{2}}{\left.{d}{x}\right.}=\frac{{{x}}^{{3}}}{{3}}$

$\displaystyle\int{2}{\left.{d}{x}\right.}={2}{x}$

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}\right|}$

We use u-substitution on int $\displaystyle\frac{{x}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$ by letting $\displaystyle{u}={{x}}^{{2}}+{2}$

then $\displaystyle{d}{u}={2}{x}\cdot{\left.{d}{x}\right.}$ rearrange into $\displaystyle{x}\cdot{\left.{d}{x}\right.}=\frac{{{d}{u}}}{{2}}$

Substituting u =x^2+2 and x * dx = (du)/2, the integral becomes:

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{u}}\cdot\frac{{{d}{u}}}{{2}}$

$\displaystyle=\frac{{1}}{{2}}\int\frac{{{d}{u}}}{{u}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{u}\right|}$

Substitute $\displaystyle{u}={{x}}^{{2}}+{2}$ then $\displaystyle\int\frac{{1}}{{2}}{\ln}{\left|{u}\right|}=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}\right|}$

$\displaystyle\int{{x}}^{{2}}{\left.{d}{x}\right.}-\int{2}{\left.{d}{x}\right.}+\int\frac{{x}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}=\frac{{{x}}^{{3}}}{{3}}-{2}{x}+\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{2}\right|}+{C}$

Ninenotes

Tuesday, April 20, 2010

$\displaystyle\int\frac{{{{x}}^{{4}}+{x}-{4}}}{{{{x}}^{{2}}+{2}}}{\left.{d}{x}\right.}$ Find the indefinite integral.

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