Question

Write the antiderivative of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\].

Answer 1

Hint: We know that the anti-derivative of a function f(x) is defined as the integral of the function f(x). So, it is clear that the anti-derivative of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\] is equal to the integral of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\]. Let us assume f(x) is equal to \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\]. Now we should find the integral of f(x). We know that \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\]. By using this concept, we can find the antiderivative of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\].

Complete step by step answer:
Before solving the problem, we should know that the anti-derivative of a function f(x) is defined as the integral of the function f(x).
So, it is clear that the anti-derivative of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\] is equal to the integral of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\].
Let us assume
\[f(x)=\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)....(1)\]
So, now we should find the integral of f(x).
\[\begin{align}
  & \Rightarrow \int{f(x)dx}=\int{\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)dx} \\
 & \Rightarrow \int{f(x)dx}=\int{3\sqrt{x}dx+\int{\dfrac{1}{\sqrt{x}}dx}}....(2) \\
\end{align}\]
Let us assume A is equal to \[\int{3\sqrt{x}dx}\].
\[\Rightarrow A=\int{3\sqrt{x}dx}....(3)\]
Let us assume B is equal to \[\int{\dfrac{1}{\sqrt{x}}dx}\].
\[\Rightarrow B=\int{\dfrac{1}{\sqrt{x}}dx}.......(4)\]
Now let us substitute equation (3) and equation (4) in equation (2).
\[\Rightarrow \int{f(x)dx}=A+B.....(5)\]
We know that \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\]. By using this formula, we should find the value of A.
\[\begin{align}
  & \Rightarrow A=\int{3\sqrt{x}dx} \\
 & \Rightarrow A=3\int{\sqrt{x}dx} \\
 & \Rightarrow A=3\int{{{x}^{\dfrac{1}{2}}}dx} \\
 & \Rightarrow A=3\left( \dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right) \\
 & \Rightarrow A=3\left( \dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right) \\
 & \Rightarrow A=3\left( \dfrac{2}{3} \right){{x}^{\dfrac{3}{2}}} \\
 & \Rightarrow A=2{{x}^{\dfrac{3}{2}}}.....(6) \\
\end{align}\]
We should also find the value of B.
\[\begin{align}
  & \Rightarrow B=\int{\dfrac{1}{\sqrt{x}}dx} \\
 & \Rightarrow B=\int{{{x}^{\dfrac{-1}{2}}}dx} \\
 & \Rightarrow B=\dfrac{{{x}^{\dfrac{-1}{2}+1}}}{\dfrac{-1}{2}+1} \\
 & \Rightarrow B=\dfrac{{{x}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \\
 & \Rightarrow B=2{{x}^{\dfrac{1}{2}}}.......(7) \\
\end{align}\]
Now let us substitute equation (6) and equation (7) in equation (5), then we get
\[\begin{align}
  & \Rightarrow \int{f(x)dx}=2{{x}^{\dfrac{3}{2}}}+2{{x}^{\dfrac{1}{2}}} \\
 & \Rightarrow \int{f(x)dx}=2{{x}^{\dfrac{1}{2}}}\left( x+1 \right).....(8) \\
\end{align}\]
From equation (8), it is clear that the value of anti-derivative of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\] is equal to \[2{{x}^{\dfrac{1}{2}}}\left( x+1 \right)\].

Note:
Some students may have a misconception that the anti-derivative of a function f(x) is equal to \[\dfrac{d}{dx}f(x)\]. If this misconception is followed, then we cannot get the correct value of anti-derivative of \[\left( 3\sqrt{x}+\dfrac{1}{\sqrt{x}} \right)\]. So, it is clear that this misconception should be avoided by students. This misconception will lead to the wrong solution.