Blog: `int x ln(1 + x) dx` First make a substitution and then use integration by parts to evaluate the integral

Tuesday, January 12, 2010

$\displaystyle\int{x}{\ln{{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}$ First make a substitution and then use integration by parts to evaluate the integral

You need to use the substitution $\displaystyle{1}+{x}={t}$ , such that:

$\displaystyle{1}+{x}={t}\Rightarrow{\left.{d}{x}\right.}={\left.{d}{t}\right.}$

Changing the variable yields:

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

You need to use the integration by parts for $\displaystyle\int{t}\cdot{\ln{{t}}}{\left.{d}{t}\right.}$ such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={\ln{{t}}}\Rightarrow{d}{u}=\frac{{1}}{{t}}$

$\displaystyle{d}{v}={t}\Rightarrow{v}=\frac{{{t}}^{{2}}}{{2}}$

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

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

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

You need to use the integration by parts for $\displaystyle\int{\ln{{t}}}{\left.{d}{t}\right.}$ such that:

$\displaystyle{u}={\ln{{t}}}\Rightarrow{d}{u}=\frac{{1}}{{t}}$

$\displaystyle{d}{v}={1}\Rightarrow{v}={t}$

$\displaystyle\int{\ln{{t}}}{\left.{d}{t}\right.}={t}\cdot{\ln{{t}}}-\int{t}\cdot\frac{{1}}{{t}}{\left.{d}{t}\right.}$

$\displaystyle\int{\ln{{t}}}{\left.{d}{t}\right.}={t}\cdot{\ln{{t}}}-{t}+{c}$

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

Replacing back the variable, yields:

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

Hence, evaluating the integral, using integration by parts, yields $\displaystyle\int{x}\cdot{\ln{{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}=\frac{{{{\left({1}+{x}\right)}}^{{2}}}}{{2}}\cdot{\ln{{\left({1}+{x}\right)}}}-\frac{{{{\left({1}+{x}\right)}}^{{2}}}}{{4}}-{\left({1}+{x}\right)}\cdot{\ln{{\left({1}+{x}\right)}}}+{\left({1}+{x}\right)}+{c}.$

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Tuesday, January 12, 2010

$\displaystyle\int{x}{\ln{{\left({1}+{x}\right)}}}{\left.{d}{x}\right.}$ First make a substitution and then use integration by parts to evaluate the integral

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