Question

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

Answer 1

Hint : As we know that integration is the process of finding functions whose derivative is given and named anti-differentiation or integration. The function is called the anti-derivative or integral or primitive of a given function $f(x)$ and $C$ is known as the constant of integration or the arbitrary constant. Here we have to integrate it by parts i.e. by using the method of completing the square in the denominator and then integrating using $u$ -substitution by the derivative of “arctan( $x$ )

Complete step by step solution:
Here we have to integrate $x$ In $(x + 1)dx$ , so e can write it as $\int {xln(x + 1)dx} = \int {\dfrac{d}{{dx}}} (\dfrac{{{x^2}}}{2})ln(x + 1)dx$
Now integrating by parts: $\int {uv' = uv - \int {u'v} }$ , It gives
We find u, v and then substitute them

$$u = \ln (x + 1) \\\ du = \dfrac{1}{{x + 1}}dx \\\ dv = x \\\ v = \dfrac{{{x^2}}}{2} \;$$

Now, we substitute them in by parts formula
$\dfrac{{{x^2}}}{2}ln(x + 1) - \int {\dfrac{{{x^2}}}{2}} \dfrac{d}{{dx}}\\{ ln(x + 1)\\} dx = \dfrac{{{x^2}}}{2}ln(x + 1) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}} dx$.
By integrating in terms of $u$ , we have $u = x + 1,du = dx$ , So by putting the values:
$\int {\dfrac{{{{(u - 1)}^2}}}{u}} du = \int {u - 2 + \dfrac{1}{u}} du$ ,
Further solving, $\dfrac{{{u^2}}}{2} - 2u + In$ $u$ $= \dfrac{{{{(x + 1)}^2}}}{2} - 2(x + 1) + ln(x + 1)$ ,
So it becomes
$= \dfrac{{{x^2}}}{2}ln(x + 1) - \dfrac{1}{2}[\dfrac{{{{(x + 1)}^2}}}{2} - 2(x + 1) + ln(x + 1)] + C \\\ \Rightarrow \dfrac{{{x^2}}}{2}ln(x + 1) - \dfrac{1}{4}{(x + 1)^2} + (x + 1) - \dfrac{1}{2}ln(x + 1) + C$ $= \dfrac{{{x^2} - 1}}{2}ln(x + 1) - \dfrac{1}{4}({x^2} + 2x + 1 - 4x - 4) + C \\\ \Rightarrow \dfrac{{{x^2} - 1}}{2}ln(x + 1) - \dfrac{1}{4}({x^2} - 2x - 3) + C$
Further solving into simpler form we have:
$= \dfrac{{{x^2} - 1}}{2}\ln (x + 1) - \dfrac{1}{4}({x^2} - 2x) + C \Rightarrow \dfrac{{{x^2} - 1}}{2}ln(x + 1) - \dfrac{1}{4}x(x - 2) + C$ .
Hence the required answer is $\dfrac{{{x^2} - 1}}{2}ln(x + 1) - \dfrac{1}{4}x(x - 2) + C$ .
So, the correct answer is “ $\dfrac{{{x^2} - 1}}{2}ln(x + 1) - \dfrac{1}{4}x(x - 2) + C$ ”.

Note : We know that integration by parts with $u = \arctan (x)$ and $dvdx = 1$ gives $v = x$ . Here we should divide in the above question whether the solution needs u or v, and try the u-substitution and then if cannot be simplified more then integrate it by parts. And the final answer must be written in the original variable of integration. It should always have $C$ , known as the constant of integration or arbitrary constant. We should always add $+ C$ as the end of the solution. The function $f(x)$ is called the integrand and $f(x)dx$ is known as the element of integration. We know that in calculus integration by substitution is also called a reverse chain rule or U substitution method.