`int (x^3 - 3x^2 + 5)/(x - 3) dx` Find the indefinite integral.

Tuesday, December 15, 2009

$\displaystyle\int\frac{{{{x}}^{{3}}-{3}{{x}}^{{2}}+{5}}}{{{x}-{3}}}{\left.{d}{x}\right.}$ Find the indefinite integral.

$\displaystyle\int\frac{{{{x}}^{{3}}-{3}{{x}}^{{2}}+{5}}}{{{x}-{3}}}{\left.{d}{x}\right.}$

To solve, divide the numerator by the denominator.

$\displaystyle=\int{\left({{x}}^{{2}}+\frac{{5}}{{{x}-{3}}}\right)}{\left.{d}{x}\right.}$

Express it as sum of two integrals.

$\displaystyle=\int{{x}}^{{2}}{\left.{d}{x}\right.}+\int\frac{{5}}{{{x}-{3}}}{\left.{d}{x}\right.}$

For the first integral, apply the formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle=\frac{{{x}}^{{3}}}{{3}}+{C}+\int\frac{{5}}{{{x}-{3}}}{\left.{d}{x}\right.}$

For the second integral, apply u-substitution method.

Let

$\displaystyle{u}={x}-{3}$

Differentiate the u.

$\displaystyle{d}{u}={\left.{d}{x}\right.}$

Then, plug-in them to the second integral.

$\displaystyle=\frac{{{x}}^{{3}}}{{3}}+{C}+{5}\int\frac{{1}}{{{x}-{3}}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{3}}}{{3}}+{C}+{5}\int\frac{{1}}{{u}}{d}{u}$

Apply the integral formula $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle=\frac{{{x}}^{{3}}}{{3}}+{5}{\ln}{\left|{u}\right|}+{C}$

And, substitute back $\displaystyle{u}={x}-{3}$ .

$\displaystyle=\frac{{{x}}^{{3}}}{{3}}+{5}{\ln}{\left|{x}-{3}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{{{x}}^{{3}}-{3}{{x}}^{{2}}+{5}}}{{{x}-{3}}}{\left.{d}{x}\right.}=\frac{{{x}}^{{3}}}{{3}}+{5}{\ln}{\left|{x}-{3}\right|}+{C}$ .

Ninenotes

Tuesday, December 15, 2009

$\displaystyle\int\frac{{{{x}}^{{3}}-{3}{{x}}^{{2}}+{5}}}{{{x}-{3}}}{\left.{d}{x}\right.}$ Find the indefinite integral.

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