Question

What is the indefinite integral of $\ln (1 + x)$ ?

Answer 1

Hint : We have to find the indefinite integral of the given function. The given differential is an indefinite one which means that we have to integrate without the boundary conditions unlike the definite integration in which we integrate with the boundary conditions. The given function is not a simple one we have to differentiate it by using the method of integration by parts. We will use the acronym ILATE when deciding the first function since $\ln (x + 1)$ is present here we will take it as the first function . The formula for integration by parts is given by,
$\int udv = uv - \int vdu$
And we will get our desired answer. The first function which is chosen should be easy to differentiate notwithstanding other questions.

Complete step-by-step answer :
The given question is $I = \int \ln \left( {1 + x} \right)dx$
Using the method of differentiation by parts which is given by ,
$\int udv = uv - \int vdu$
For this $\ln (x + 1)$ is the first function .
For $\int \ln \left( {1 + x} \right)dx$

{u = \ln \left( {1 + x} \right)\;} \\\ {dv = dx \Rightarrow v = x} \end{array}$$ Also $ du = \dfrac{1}{{1 + x}}dx $ Upon solving using the formula by taking these values we can write, $ I = x\ln \left( {1 + x} \right) - \int \dfrac{x}{{1 + x}}dx $ From here for integrating the second function we write as, $ I = x\ln \left( {1 + x} \right) - \int \dfrac{{1 + x - 1}}{{1 + x}}dx $ $ I = x\ln \left( {1 + x} \right) - \int \left( {\dfrac{{1 + x}}{{1 + x}} - \dfrac{1}{{1 + x}}} \right)dx $ Rearranging the integral we get, $ I = x\ln \left( {1 + x} \right) - \int \left( {1 - \dfrac{1}{{1 + x}}} \right)dx $ Upon integrating we get, $$I = x\ln \left( {1 + x} \right) - x + \ln \left( {1 + x} \right) + K$$ Elegantly rearranging we get, $ I = \left( {x + 1} \right)\ln \left( {1 + x} \right) - x + K $ Thus we get our final answer. **So, the correct answer is “ $ I = \left( {x + 1} \right)\ln \left( {1 + x} \right) - x + K $ ”.** **Note** : When we do the indefinite integration we always have to remember to put the constant of integration at the end. This is mandatory and omitting it will result in unfortunate deduction of marks. But whenever we do definite integration ie. When we integrate with the boundary conditions we do not need to write it because it gets cancelled due to limits.