Question

How do you integrate \[{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)\]?

Answer 1

Hint: This type of question is based on the concept of integration by parts. Since the given function can be divided into two parts, we have to use integration by parts with respect to t, that is \[\int{udv=uv-\int{vdu}}\]. Here, we find that u= ln(t) and \[dv={{t}^{\dfrac{1}{2}}}dt\]. Integrate dv to find v and differentiate u to find du, with respect to t. then substitute in the formula of integration by parts. Using the power rule of integration that is \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\], solve the indefinite integration.

Complete step by step solution:
According to the question, we are asked to find the integral of \[{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)\].
We have been given the function is \[{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)\]. --------(1)
We find that the given function is with respect to t.
That is, we have to find \[\int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}\].
To integrate function (1), we have to use integration by parts.
We know that the formula for integration by parts is \[\int{udv=uv-\int{vdu}}\].
Here, let us consider u=ln(t) and \[dv={{t}^{\dfrac{1}{2}}}dt\].
We have to integrate dv with respect to t to find the value of v.
\[\Rightarrow \int{dv}=\int{\left( {{t}^{\dfrac{1}{2}}} \right)}dt\]
Using the power rule of integration that is \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\], we get
\[\int{dv}=\dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}\]
We know that \[\int{dv=v}\]. Therefore, we get
\[v=\dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}\]
On taking LCM in the power of the numerator and term in the denominator, we get
\[v=\dfrac{{{t}^{\dfrac{1+2}{2}}}}{\dfrac{1+2}{2}}\]
\[\Rightarrow v=\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}\]
Using the rule \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\], we get
\[v=\dfrac{2{{t}^{\dfrac{3}{2}}}}{3}\]
Now, consider u=ln(t).
Differentiate u with respect to t.
\[\dfrac{du}{dt}=\dfrac{d}{dt}\left( \ln \left( t \right) \right)\]
We know that \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\]. Using this rule of integration, we get
\[\dfrac{du}{dt}=\dfrac{1}{t}\]
Now, take the dt from the denominator of the LHS to the RHS, we get
\[\Rightarrow du=\dfrac{1}{t}dt\].
Now, substitute the value of u, v, du and dv in the integration by parts formula, that is \[\int{udv=uv-\int{vdu}}\].
\[\Rightarrow \int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\ln \left( t \right)\times \dfrac{2{{t}^{\dfrac{3}{2}}}}{3}-\int{\dfrac{2{{t}^{\dfrac{3}{2}}}}{3}\times \dfrac{1}{t}dt}+c\]
\[\Rightarrow \int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3}\ln \left( t \right){{t}^{\dfrac{3}{2}}}-\int{\dfrac{2{{t}^{\dfrac{3}{2}}}}{3}\times \dfrac{1}{t}dt}+c\]
Taking the constants out of the integration, we get
\[\int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3}\ln \left( t \right){{t}^{\dfrac{3}{2}}}-\dfrac{2}{3}\int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}+c\] -------------(2)
Let us now consider \[\int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}\].
We know that \[\dfrac{1}{x}={{x}^{-1}}\]. Using this rule in the integration, we get
\[\Rightarrow \int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\int{{{t}^{\dfrac{3}{2}}}\times {{t}^{-1}}dt}\]
Using the property of powers, that is \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], we get
\[\int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\int{{{t}^{\dfrac{3}{2}-1}}dt}\]
Now, we need to take the LCM in the power.
\[\Rightarrow \int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\int{{{t}^{\dfrac{3-2}{2}}}dt}\]
\[\Rightarrow \int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\int{{{t}^{\dfrac{1}{2}}}dt}\]
Using the power rule of integration that is \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\], we get
\[\int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}\]
On taking LCM, we get
\[\int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\dfrac{{{t}^{\dfrac{1+2}{2}}}}{\dfrac{1+2}{2}}\]
\[\Rightarrow \int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}\]
Using the rule \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\], we get
\[\Rightarrow \int{{{t}^{\dfrac{3}{2}}}\times \dfrac{1}{t}dt}=\dfrac{2{{t}^{\dfrac{3}{2}}}}{3}\]
Substitute the value in equation (2).
\[\Rightarrow \int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3}\ln \left( t \right){{t}^{\dfrac{3}{2}}}-\dfrac{2}{3}\times \dfrac{2{{t}^{\dfrac{3}{2}}}}{3}+c\]
Let us now take \[\dfrac{2}{3}\] from the RHS. We get
\[\int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3}\left[ \ln \left( t \right){{t}^{\dfrac{3}{2}}}-\dfrac{2{{t}^{\dfrac{3}{2}}}}{3} \right]+c\].
We can also take \[{{t}^{\dfrac{3}{2}}}\] common from the RHS. We get
\[\int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\left[ \ln \left( t \right)-\dfrac{2}{3} \right]+c\]
Therefore, the integration of \[{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)\] is \[\dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\left[ \ln \left( t \right)-\dfrac{2}{3} \right]+c\].

Note: We can further simplify the final answer obtained.
Take 3 as the LCM in the terms inside the bracket.
\[\int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\left[ \dfrac{3\ln \left( t \right)-2}{3} \right]+c\]
Now, take 3 outside the bracket from the denominator.
\[\Rightarrow \int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{3\times 3}{{t}^{\dfrac{3}{2}}}\left[ 3\ln \left( t \right)-2 \right]+c\]
On further simplifications, we get
\[\int{{{t}^{\dfrac{1}{2}}}\cdot \ln \left( t \right)dt}=\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}\left[ 3\ln \left( t \right)-2 \right]+c\].