Question

How do you integrate \[\int{\ln \left( x+1 \right)}\] by integration by parts methods?

Answer 1

Hint: For solving this question, we are going to use integration by parts formula. First, we will understand what integration by parts formula is. After that, we apply that formula in the question and find the way to the solution. The formula we are going to apply in this question is: \[\int{udv=}uv-\int{vdu}\]

Complete step by step answer:
Let us solve the question.
First, we will understand what integration by parts formula is and how to apply it.
Integration by parts formula is used to transform the anti-derivative or integral of a product of two functions for which solutions can be easily found.
Let the two functions be u(x) and v(x).
Then, according to integration by parts formula,
\[\int{udv=}uv-\int{vdu}\]
The integral of the product of the two functions is taken by considering the left term that is, u as the first function and second term that is, v as the second function. The first function u is chosen in such a way that the derivative of the function could be easily integrated. Whenever we use integration by parts formula, an ILATE Rule is followed. This rule says that it will take u as the first function according to the following order:
I stands for Inverse trigonometric function
L stands Logarithmic functions
A stands for Algebraic functions
T stands for Trigonometric functions and
E stands for exponential functions.
We have to find the integration of \[\int{\ln \left( x+1 \right)}dx\] .
Now, here in the question, from the above formula of integration by parts, we can say that \[\begin{align}
  & dv=dx\text{ }\Rightarrow \text{ }v=x\text{ and} \\
 & u=\ln (x+1)\text{ }\Rightarrow \text{ }du=\dfrac{1}{x+1}dx \\
\end{align}\]
Here, we have taken ln(x+1) as the first function because logarithmic functions come before algebraic functions, that ln(x+1) comes before 1.
Now, using the formula integration by parts, we can write
\[\begin{align}
  & I=\int{\ln (x+1)dx=}x\ln (x+1)-\int{\left( \dfrac{x}{1+x} \right)}dx \\
 & \Rightarrow I=x\ln (x+1)-\int{\left( \dfrac{1+x-1}{1+x} \right)dx} \\
 & \Rightarrow I=x\ln (x+1)-\int{\left( \dfrac{1+x}{1+x}-\frac{1}{1+x} \right)}dx \\
 & \Rightarrow I=x\ln (x+1)-\int{\left( \left( 1 \right)-\left( \dfrac{1}{1+x} \right) \right)}dx \\
 & \Rightarrow I=x\ln (x+1)-\left( \int{1}dx-\int{\dfrac{1}{1+x}dx} \right) \\
\end{align}\]
Both of these are simple integrals:
\[\begin{align}
  & \Rightarrow I=x\ln (x+1)-\left( x-\ln (1+x) \right)+C \\
 & \Rightarrow I=x\ln (x+1)-x+\ln (1+x)+C \\
\end{align}\]
Factoring ln(x+1), we get
\[\Rightarrow I=(x+1)\ln (x+1)-x+C\]

Note: In this type of question, we should have a proper knowledge in integration and integration by parts for solving. To use this formula of integration by parts, we need to identify u and v. For making the solving process of the question easier, we should use the ILATE Rule. It helps too much in the integration by parts during integration.