Question

How do you find the indefinite integral of $ \sqrt{25+{{x}^{2}}} $ ?

Answer 1

Hint: We have to find the indefinite integral of $ \sqrt{25+{{x}^{2}}} $ . The given integral is in the form of $ \sqrt{{{a}^{2}}+{{x}^{2}}} $ where the indefinite integral gives us $ \int{\sqrt{{{a}^{2}}+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+c $ . We find the value of $ a $ from the relation of $ {{a}^{2}}=25 $ . Then we place the value of $ a $ in the integral to find the solution of the integral.

Complete step by step answer:
We need to find the indefinite integral of the given function $ \sqrt{25+{{x}^{2}}} $ .
It is of the form $ \sqrt{{{a}^{2}}+{{x}^{2}}} $ where $ a $ is a constant.
We know that the indefinite integral of the $ \sqrt{{{a}^{2}}+{{x}^{2}}} $ is
 $ \int{\sqrt{{{a}^{2}}+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+c $ .
For our given case the value of $ a $ will be decided from the relation of $ {{a}^{2}}=25 $ .
We get the value of $ a $ as $ a=\sqrt{25}=5 $ .
We apply the value of $ a=5 $ in the integral form of $ \int{\sqrt{{{a}^{2}}+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+c $ to get
 $ \int{\sqrt{25+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{25+{{x}^{2}}}+\dfrac{25}{2}\log \left| x+\sqrt{25+{{x}^{2}}} \right|+c $ .
Therefore, the indefinite integral value of $ \sqrt{25+{{x}^{2}}} $ is $ \int{\sqrt{25+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{25+{{x}^{2}}}+\dfrac{25}{2}\log \left| x+\sqrt{25+{{x}^{2}}} \right|+c $ .

Note:
 The integral form $ \int{\sqrt{{{a}^{2}}+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+c $ is a derivation of the use of the by parts theorem. If the theorem is hard to remember then we solve the integral of $ \sqrt{25+{{x}^{2}}} $ by parts.
We take two function in the integral formula $ \int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx} $ where
 $ u=1,v=\sqrt{25+{{x}^{2}}} $ .
We place the functions and get
 $ \int{\left( 1\times \sqrt{25+{{x}^{2}}} \right)dx}=\sqrt{25+{{x}^{2}}}\int{dx}-\int{\left[ \dfrac{d}{dx}\left( \sqrt{25+{{x}^{2}}} \right)\int{dx} \right]dx} $ .
The differentiation of $ \sqrt{25+{{x}^{2}}} $ will be $ \dfrac{d}{dx}\left( \sqrt{25+{{x}^{2}}} \right)=\dfrac{2x}{2\sqrt{25+{{x}^{2}}}}=\dfrac{x}{\sqrt{25+{{x}^{2}}}} $ .

Therefore, $ I=\int{\sqrt{25+{{x}^{2}}}dx}=x\sqrt{25+{{x}^{2}}}-\int{\dfrac{{{x}^{2}}}{\sqrt{25+{{x}^{2}}}}dx} $ .
We can change the numerator $ \dfrac{{{x}^{2}}}{\sqrt{25+{{x}^{2}}}}=\dfrac{{{x}^{2}}+25-25}{\sqrt{25+{{x}^{2}}}}=\dfrac{{{x}^{2}}+25}{\sqrt{25+{{x}^{2}}}}-\dfrac{25}{\sqrt{25+{{x}^{2}}}} $ .
Integrating we get $ \int{\dfrac{{{x}^{2}}}{\sqrt{25+{{x}^{2}}}}dx}=\int{\dfrac{{{x}^{2}}+25}{\sqrt{25+{{x}^{2}}}}dx}-25\int{\dfrac{25}{\sqrt{25+{{x}^{2}}}}dx} $ .
We replace the integral value. It gives
 $ \begin{align}
  & \int{\dfrac{{{x}^{2}}}{\sqrt{25+{{x}^{2}}}}dx}=\int{\sqrt{25+{{x}^{2}}}dx}-25\int{\dfrac{25}{\sqrt{25+{{x}^{2}}}}dx} \\
 & \Rightarrow \int{\dfrac{{{x}^{2}}}{\sqrt{25+{{x}^{2}}}}dx}=I-25\log \left| x+\sqrt{25+{{x}^{2}}} \right| \\
\end{align} $
We again replace the values and get
 $ \begin{align}
  & I=x\sqrt{25+{{x}^{2}}}-\int{\dfrac{{{x}^{2}}}{\sqrt{25+{{x}^{2}}}}dx} \\
 & \Rightarrow I=x\sqrt{25+{{x}^{2}}}-\left[ I-25\log \left| x+\sqrt{25+{{x}^{2}}} \right| \right] \\
 & \Rightarrow 2I=x\sqrt{25+{{x}^{2}}}+25\log \left| x+\sqrt{25+{{x}^{2}}} \right| \\
 & \Rightarrow I=\dfrac{x}{2}\sqrt{25+{{x}^{2}}}+\dfrac{25}{2}\log \left| x+\sqrt{25+{{x}^{2}}} \right|+c \\
\end{align} $
Thus, verified the indefinite integral value of $ \sqrt{25+{{x}^{2}}} $ is $ \int{\sqrt{25+{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{25+{{x}^{2}}}+\dfrac{25}{2}\log \left| x+\sqrt{25+{{x}^{2}}} \right|+c $ .