New Notes Blog: `int tln(t+1) dt` Find the indefinite integral

Sunday, 30 November 2014

$\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}$ Find the indefinite integral

Recall that indefinite integral follows $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$ where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as the constant of integration.

For the given integral problem: $\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}$ , we may apply u-substitution by letting:

$\displaystyle{u}={t}+{1}$ that can be rearrange as $\displaystyle{t}={u}-{1}$ .

The derivative of u is $\displaystyle{d}{u}={\left.{d}{t}\right.}$ .

Plug-in the values, we get:

$\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}=\int{\left({u}-{1}\right)}{\ln{{\left({u}\right)}}}\ldots$

Recall that indefinite integral follows $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$ where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as the constant of integration.

For the given integral problem: $\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}$ , we may apply u-substitution by letting:

$\displaystyle{u}={t}+{1}$ that can be rearrange as $\displaystyle{t}={u}-{1}$ .

The derivative of u is $\displaystyle{d}{u}={\left.{d}{t}\right.}$ .

Plug-in the values, we get:

$\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}=\int{\left({u}-{1}\right)}{\ln{{\left({u}\right)}}}{d}{u}$

Apply integration by parts: $\displaystyle\int{f{\cdot}}{g{'}}={f{\cdot}}{g{-}}\int{g{\cdot}}{f{'}}$ .

We may let:

$\displaystyle{f{=}}{\ln{{\left({u}\right)}}}$ then $\displaystyle{f{'}}=\frac{{{d}{u}}}{{u}}$

$\displaystyle{g{'}}={u}-{1}{d}{u}$ then $\displaystyle{g{=}}\frac{{{u}}^{{2}}}{{2}}-{u}$

Note: $\displaystyle{g{=}}\int{g{'}}=\int{\left({u}+{1}\right)}{d}{u}$ .

$\displaystyle\int{\left({u}-{1}\right)}{d}{u}=\int{\left({u}\right)}{d}{u}-\int{\left({1}\right)}{d}{u}$

$\displaystyle=\frac{{{u}}^{{{1}+{1}}}}{{{1}+{1}}}-{1}{u}$

$\displaystyle=\frac{{{u}}^{{2}}}{{2}}-{u}$

Applying the formula for integration by parts, we set it up as:

$\displaystyle\int{\left({u}-{1}\right)}{\ln{{\left({u}\right)}}}{d}{u}={\ln{{\left({u}\right)}}}\cdot{\left(\frac{{{u}}^{{2}}}{{2}}-{u}\right)}-\int{\left(\frac{{{u}}^{{2}}}{{2}}-{u}\right)}\cdot\frac{{{d}{u}}}{{u}}$

$\displaystyle=\frac{{{{u}}^{{2}}{\ln{{\left({u}\right)}}}}}{{2}}-{u}\cdot{\ln{{\left({u}\right)}}}-\int{\left(\frac{{{u}}^{{2}}}{{{2}{u}}}-\frac{{u}}{{u}}\right)}{d}{u}$

$\displaystyle=\frac{{{{u}}^{{2}}{\ln{{\left({u}\right)}}}}}{{2}}-{u}\cdot{\ln{{\left({u}\right)}}}-\int{\left(\frac{{u}}{{2}}-{1}\right)}{d}{u}$

For the integral part: $\displaystyle\int{\left(\frac{{u}}{{2}}-{1}\right)}{d}{u}$ , we apply the basic integration property: $\displaystyle\int{\left({u}-{v}\right)}{\left.{d}{x}\right.}=\int{\left({u}\right)}{\left.{d}{x}\right.}-\int{\left({v}\right)}{\left.{d}{x}\right.}$ .

$\displaystyle\int{\left(\frac{{u}}{{2}}-{1}\right)}{d}{u}=\int{\left(\frac{{u}}{{2}}\right)}{d}{u}-\int{\left({1}\right)}{d}{u}$

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

$\displaystyle=\frac{{1}}{{2}}\cdot{\left(\frac{{{u}}^{{2}}}{{2}}\right)}-{1}\cdot{u}+{C}$

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

Applying $\displaystyle\int{\left(\frac{{u}}{{2}}-{1}\right)}{d}{u}=\frac{{{u}}^{{2}}}{{4}}-{u}+{C}$ , we get:

$\displaystyle\int{\left({u}-{1}\right)}{\ln{{\left({u}\right)}}}{d}{u}=\frac{{{{u}}^{{2}}{\ln{{\left({u}\right)}}}}}{{2}}-{\underline{{n}}}{\left({u}\right)}-\int{\left(\frac{{u}}{{2}}-{1}\right)}{d}{u}$

$\displaystyle=\frac{{{{u}}^{{2}}{\ln{{\left({u}\right)}}}}}{{2}}-{u}\cdot{\ln{{\left({u}\right)}}}-{\left[\frac{{{u}}^{{2}}}{{4}}-{u}\right]}+{C}$

$\displaystyle=\frac{{{{u}}^{{2}}{\ln{{\left({u}\right)}}}}}{{2}}-{u}\cdot{\ln{{\left({u}\right)}}}-\frac{{{u}}^{{2}}}{{4}}+{u}+{C}$

Plug-in $\displaystyle{u}={t}+{1}$ on $\displaystyle\frac{{{{u}}^{{2}}{\ln{{\left({u}\right)}}}}}{{2}}-{u}\cdot{\ln{{\left({u}\right)}}}-\frac{{{u}}^{{2}}}{{4}}+{u}+{C}$ , we get the complete indefinite integral as:

$\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}=\frac{{{{\left({t}+{1}\right)}}^{{2}}{\ln{{\left({t}+{1}\right)}}}}}{{2}}-{\left({t}+{1}\right)}{\ln{{\left({t}+{1}\right)}}}-\frac{{{\left({t}+{1}\right)}}^{{2}}}{{4}}+{t}+{1}+{C}$

OR $\displaystyle{\left[\frac{{{\left({t}+{1}\right)}}^{{2}}}{{2}}-{t}-{1}\right]}{\ln{{\left({t}+{1}\right)}}}-\frac{{{\left({t}+{1}\right)}}^{{2}}}{{4}}+{t}+{1}+{C}$

New Notes Blog

Sunday, 30 November 2014

$\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}$ 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, 30 November 2014

Find the indefinite integral

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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?

$\displaystyle\int{t}{\ln{{\left({t}+{1}\right)}}}{\left.{d}{t}\right.}$ 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?