New Notes Blog: `int 1/(x^2-4x+9) dx` Find the indefinite integral

Sunday, 19 January 2014

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}$ Find the indefinite integral

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}$

Let's complete the square for the denominator of the integral as:

$\displaystyle{\left({{x}}^{{2}}-{4}{x}+{9}\right)}={{\left({x}-{2}\right)}}^{{2}}+{5}$

$\displaystyle{{\left({x}-{2}\right)}}^{{2}}+{{\left(\sqrt{{{5}}}\right)}}^{{2}}$

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{{\left({x}-{2}\right)}}^{{2}}+{{\left(\sqrt{{{5}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

Let's apply the integral substitution,

substitute $\displaystyle{u}={x}-{2}$

$\displaystyle{d}{u}={1}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{u}}^{{2}}+{{\left(\sqrt{{{5}}}\right)}}^{{2}}}}{d}{u}$

Now use the standard integral : $\displaystyle\int\frac{{1}}{{{{x}}^{{2}}+{{a}}^{{2}}}}=\frac{{1}}{{a}}{\arctan{{\left(\frac{{x}}{{a}}\right)}}}$

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}{\arctan{{\left(\frac{{u}}{\sqrt{{{5}}}}\right)}}}$

substitute back u=(x-2) and add a constant C to the solution,

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}{\arctan{{\left(\frac{{{x}-{2}}}{\sqrt{{{5}}}}\right)}}}+{C}$

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}$

Let's complete the square for the denominator of the integral as:

$\displaystyle{\left({{x}}^{{2}}-{4}{x}+{9}\right)}={{\left({x}-{2}\right)}}^{{2}}+{5}$

$\displaystyle{{\left({x}-{2}\right)}}^{{2}}+{{\left(\sqrt{{{5}}}\right)}}^{{2}}$

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{{\left({x}-{2}\right)}}^{{2}}+{{\left(\sqrt{{{5}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

Let's apply the integral substitution,

substitute $\displaystyle{u}={x}-{2}$

$\displaystyle{d}{u}={1}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{u}}^{{2}}+{{\left(\sqrt{{{5}}}\right)}}^{{2}}}}{d}{u}$

Now use the standard integral : $\displaystyle\int\frac{{1}}{{{{x}}^{{2}}+{{a}}^{{2}}}}=\frac{{1}}{{a}}{\arctan{{\left(\frac{{x}}{{a}}\right)}}}$

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}{\arctan{{\left(\frac{{u}}{\sqrt{{{5}}}}\right)}}}$

substitute back u=(x-2) and add a constant C to the solution,

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}{\arctan{{\left(\frac{{{x}-{2}}}{\sqrt{{{5}}}}\right)}}}+{C}$

New Notes Blog

Sunday, 19 January 2014

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Sunday, 19 January 2014

Find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle\int\frac{{1}}{{{{x}}^{{2}}-{4}{x}+{9}}}{\left.{d}{x}\right.}$ Find the indefinite integral

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?