Question

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

Answer 1

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).

& \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{aligned}$$ Let us assume A is equal to $$\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{aligned} & \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{aligned}$$ We should also find the value of B. $$\begin{aligned} & \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{aligned}$$ Now let us substitute equation (6) and equation (7) in equation (5), then we get $$\begin{aligned} & \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{aligned}$$ From equation (8), it is clear that the value of anti-derivative of $$\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.