Question

How do you evaluate the given indefinite integral $\int{\ln \left( t+1 \right)dt}$?

Answer 1

Hint: We start solving the problem by equating the given indefinite integral to a variable. We then recall the integration by parts as $\int{f\left( x \right)\times g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( \dfrac{d\left( f\left( x \right) \right)}{dx}\int{g\left( x \right)dx} \right)dx}$ to proceed through the problem. We then make the necessary calculations and make use of the results $\int{adx}=ax+C$, $\dfrac{d\left( \ln y \right)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}$ to proceed further through the problem. We then make use of the fact that $\int{\dfrac{1}{a+x}dx}=\ln \left( a+x \right)+C$ and make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the result of an indefinite integral $\int{\ln \left( t+1 \right)dt}$.
Let us assume $I=\int{\ln \left( t+1 \right)dt}$.
$\Rightarrow I=\int{\ln \left( t+1 \right)\times 1dt}$ ---(1).
We can see that the integrand is in the form of $\int{f\left( x \right)\times g\left( x \right)dx}$. From integration by parts, we know that $\int{f\left( x \right)\times g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( \dfrac{d\left( f\left( x \right) \right)}{dx}\int{g\left( x \right)dx} \right)dx}$. Let us use this result in equation (1).
\[\Rightarrow I=\ln \left( t+1 \right)\int{1dt}-\int{\left( \dfrac{d\left( \ln \left( t+1 \right) \right)}{dt}\int{1dt} \right)dt}\] ---(2).
We know that $\int{adx}=ax+C$ and $\dfrac{d\left( \ln y \right)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}$. Let us use these results in equation (2).
\[\Rightarrow I=t\ln \left( t+1 \right)-\int{\left( \dfrac{t}{t+1} \right)dt}\].
\[\Rightarrow I=t\ln \left( t+1 \right)-\int{\left( \dfrac{t+1-1}{t+1} \right)dt}\].
\[\Rightarrow I=t\ln \left( t+1 \right)-\int{\left( 1-\dfrac{1}{t+1} \right)dt}\].
\[\Rightarrow I=t\ln \left( t+1 \right)-\int{1dt}+\int{\dfrac{1}{t+1}dt}\] ---(4).
We know that $\int{adx}=ax+C$ and $\int{\dfrac{1}{a+x}dx}=\ln \left( a+x \right)+C$. Let us use this result in equation (4).
\[\Rightarrow I=t\ln \left( t+1 \right)-t+\ln \left( t+1 \right)+C\].
\[\Rightarrow I=\left( t+1 \right)\ln \left( t+1 \right)-t+C\].

$\therefore $ We have found the result of integration $\int{\ln \left( t+1 \right)dt}$ as \[\left( t+1 \right)\ln \left( t+1 \right)-t+C\].

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not forget to add constants of integration while solving the problems related to indefinite integrals. Similarly, we can expect problems to find the value of $\int\limits_{0}^{3}{\ln \left( t+1 \right)dt}$. We can solve the problem by first assuming $1+t=x$ in the integral and then performing the integration by parts to get the required answer.