Blog: `int (x^3 - 8x)/x^2 dx` Find the indefinite integral.

$\displaystyle\int\frac{{{{x}}^{{3}}-{8}{x}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$

To solve, express the integrand as two fractions with same denominators.

$\displaystyle=\int{\left(\frac{{{x}}^{{3}}}{{{x}}^{{2}}}-\frac{{{8}{x}}}{{{x}}^{{2}}}\right)}{\left.{d}{x}\right.}$

Simplify each fraction.

$\displaystyle=\int{\left({x}-\frac{{8}}{{x}}\right)}{\left.{d}{x}\right.}$

Express it as difference of two integrals.

$\displaystyle=\int{x}{\left.{d}{x}\right.}-\int\frac{{8}}{{x}}{\left.{d}{x}\right.}$

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

And for the second integral, apply the formula $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle=\int{x}{\left.{d}{x}\right.}-{8}\int\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{2}}}{{2}}-{8}{\ln}{\left|{x}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{{{x}}^{{3}}-{8}{x}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}=\frac{{{x}}^{{2}}}{{2}}-{8}{\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle\int\frac{{{{x}}^{{3}}-{8}{x}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$

To solve, express the integrand as two fractions with same denominators.

$\displaystyle=\int{\left(\frac{{{x}}^{{3}}}{{{x}}^{{2}}}-\frac{{{8}{x}}}{{{x}}^{{2}}}\right)}{\left.{d}{x}\right.}$

Simplify each fraction.

$\displaystyle=\int{\left({x}-\frac{{8}}{{x}}\right)}{\left.{d}{x}\right.}$

Express it as difference of two integrals.

$\displaystyle=\int{x}{\left.{d}{x}\right.}-\int\frac{{8}}{{x}}{\left.{d}{x}\right.}$

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

And for the second integral, apply the formula $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle=\int{x}{\left.{d}{x}\right.}-{8}\int\frac{{1}}{{x}}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{{x}}^{{2}}}{{2}}-{8}{\ln}{\left|{x}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{{{x}}^{{3}}-{8}{x}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}=\frac{{{x}}^{{2}}}{{2}}-{8}{\ln}{\left|{x}\right|}+{C}$ .

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$\displaystyle\int\frac{{{{x}}^{{3}}-{8}{x}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$ Find the indefinite integral.

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What are the problems with Uganda's government?

Find the indefinite integral.

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What are the problems with Uganda&#39;s government?

$\displaystyle\int\frac{{{{x}}^{{3}}-{8}{x}}}{{{x}}^{{2}}}{\left.{d}{x}\right.}$ Find the indefinite integral.

What are the problems with Uganda's government?