Question

How do you integrate $x\ln \left( 1+x \right)dx?$

Answer 1

Here as you know that we have to integrate the following expression $x\ln \left( 1+x \right)dx$ Integration is the way of inverse process of differentiation. Instead of differentiating the function, you are given the derivative of a function. You are given the derivative of a function and asked to find its primitive i.e. original function. Such a process is called integration or anti-differentiation. Use integration by parts. Integration by parts is like the product rule of integration to find the product of functions in terms of the integral of the product of their derivatives of antiderivative. In fact it is derived from the product rule it states that.
$\int{x{{e}^{x}}dx=x{{e}^{x}}-\int{{{e}^{x}}dx=x{{e}^{x}}-{{e}^{x}}+c}}$

Complete step by step solution:
As per the question, $x\ln \left( 1+x \right)dx$
Let,
$u=x+1$
$\Rightarrow du=dx$
$\Rightarrow x=u-1$
Therefore the modified equation will be, and also integrate.
$\int{\left( u-1 \right)\ln udu}$
$\Rightarrow \int{u\ln udu}-\int{\ln udu}$
With these two integrals in mind, you can do integration by points consuming you ahead by know the integral of $\ln x$
The $\int{\ln u}$ let,
$\Rightarrow s=\ln u$
$\Rightarrow ds=\dfrac{1}{u}du$
$\Rightarrow dt=udu$
$\Rightarrow t=\dfrac{{{u}^{2}}}{2}$
Thus,
$\Rightarrow St=\int{tds}$
$=\left[ \dfrac{{{u}^{2}}\ln u}{2}-\int{\dfrac{{{u}^{2}}}{2}-\dfrac{1}{u}du} \right]-\int{\ln udu}$
Here, you have to put the value of $s,t$ and $ds.$
$\Rightarrow =\dfrac{{{u}^{2}}\ln u}{2}-\dfrac{1}{2}\int{udu-\int{\ln udu}}$
Here, $u$ and ${{u}^{2}}=u\times u$ get canceled from ${{u}^{2}}$ only one $u$ will get $'a'$ will remain.
Here the integration of $\int{udu}$ is $\dfrac{{{u}^{2}}}{2}$
Therefore the modified equation will be,
$=\dfrac{{{u}^{2}}\ln u}{2}-\dfrac{1}{2}.\dfrac{{{u}^{2}}}{2}-\left( u\ln u-u \right)$
$\Rightarrow =\dfrac{{{u}^{2}}\ln u}{2}-\dfrac{{{u}^{2}}}{u}-\left( u\ln -u \right)$
As after opening bracket $'-u'$ converted into $'+u'$
$\Rightarrow =\dfrac{{{u}^{2}}\ln u}{2}-\dfrac{{{u}^{2}}}{u}-u\ln u+u$
$\Rightarrow =\dfrac{{{u}^{2}}\ln u}{2}-u\ln u-\dfrac{{{u}^{2}}}{u}+u$
Replace, $x+1$ on the place of $u$
$dx$ on the place of $du$
$x$ on the place of $u-1$
Hence,
$=\dfrac{{{\left( x+1 \right)}^{2}}\ln \left( x+1 \right)}{2}-\left( x+1 \right)\ln \left( x+1 \right)-\dfrac{{{\left( x+1 \right)}^{2}}}{4}+x$
And $1$ gets embedded into $C.$
$=\dfrac{{{\left( x+1 \right)}^{2}}\ln \left( x+1 \right)}{2}-\left( x+1 \right)\ln \left( x+1 \right)-\dfrac{{{\left( x+1 \right)}^{2}}}{4}+x+C$

Additional Information:
When one has to integrate a product of two functions, integration by parts is useful.
If $f(x)=g(x).h(x)$
Then,
$\Rightarrow \int{f(x)dx=g(x)\int{h(x)dx-\int{\dfrac{d}{dx}g(x).\left( h(x)dx \right)}dx}}$
This is called integration by parts.
The integral of the product of two functions may be vertically given as.
“first function into integral of the second minus integral of the derivative of the first into integral of the second '' Which is nothing but integration by parts. Now, one thing that must be noted is that the correct choice of first function and second function can either make or break a problem. The correct choice can vastly simplify and an incorrect one can put you in a lot of trouble.
For instance,
$\Rightarrow$ $\int{x.{{e}^{x}}dx}$ is the integral you need to evaluate.
If you see, use ${{e}^{x}}$ as the first function and $x$ as the second and integrate by parts.
$\Rightarrow$ $\int{x{{e}^{x}}dx}={{e}^{x}}\int{xdx-\int{\left( {{e}^{x}}xdx \right)}dx}$
$={{e}^{x}}\dfrac{{{x}^{2}}}{2}-\int{{{e}^{x}}.\dfrac{{{x}^{2}}}{2}dx+c}$
If you apply integration by parts to the second term. You again get a term with a ${{x}^{3}}$ and so on.
This not only complicates the problem but spells disaster.
But, if you had chosen $'x'$ to be first and $'{{e}^{x}}'$ to be second the integral would have been simply to evaluate.
$\Rightarrow$ $\int{x.{{e}^{x}}dx=x\int{{{e}^{x}}dx-\int{\left( \dfrac{d}{dx}.x\int{{{e}^{x}}dx} \right)dx}}}$
$\Rightarrow$ $\int{x.{{e}^{x}}dx={{e}^{x}}x.{{e}^{x}}+c}$

Note: Integration by parts is used only where we have to find the integral product of functions in terms of integral of the product of their derivative or anti-derivative.
Apply integration by parts that is $f(x)=g(x).h(x)$
Replace the given term by $'u'$ which will get easy to solve and also save time again replacing $'u'$ by $'x'$ after getting the solution.