New Notes Blog: `int (x^2 - x^-2) dx` Find the general indefinite integral.

Sunday, 17 August 2014

$\displaystyle\int{\left({{x}}^{{2}}-{{x}}^{{-{{2}}}}\right)}{\left.{d}{x}\right.}$ Find the general indefinite integral.

You need to evaluate the indefinite integral, such that:

$\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{c}$

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

Evaluating each definite integral, using the formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{{x}}^{{{n}+{1}}}}}{{{n}+{1}}}+{c}$ , yields:

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

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

Gathering the results, yields:

$\displaystyle\int{\left({{x}}^{{2}}-{{x}}^{{-{2}}}\right)}{\left.{d}{x}\right.}=\frac{{{{x}}^{{3}}}}{{3}}-{\left(-\frac{{1}}{{x}}\right)}+{c}=\frac{{{{x}}^{{3}}}}{{3}}\ldots$

You need to evaluate the indefinite integral, such that:

$\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{c}$

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

Evaluating each definite integral, using the formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{{x}}^{{{n}+{1}}}}}{{{n}+{1}}}+{c}$ , yields:

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

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

Gathering the results, yields:

$\displaystyle\int{\left({{x}}^{{2}}-{{x}}^{{-{2}}}\right)}{\left.{d}{x}\right.}=\frac{{{{x}}^{{3}}}}{{3}}-{\left(-\frac{{1}}{{x}}\right)}+{c}=\frac{{{{x}}^{{3}}}}{{3}}+\frac{{1}}{{x}}+{c}$

Hence, evaluating the indefinite integral yields $\displaystyle\int{\left({{x}}^{{2}}-{{x}}^{{-{2}}}\right)}{\left.{d}{x}\right.}=\frac{{{{x}}^{{3}}}}{{3}}+\frac{{1}}{{x}}+{c}.$

New Notes Blog

Sunday, 17 August 2014

$\displaystyle\int{\left({{x}}^{{2}}-{{x}}^{{-{{2}}}}\right)}{\left.{d}{x}\right.}$ Find the general 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, 17 August 2014

Find the general indefinite integral.

No comments:

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?

$\displaystyle\int{\left({{x}}^{{2}}-{{x}}^{{-{{2}}}}\right)}{\left.{d}{x}\right.}$ Find the general 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?