Question

Integrate $ \log (1 + {x^2}) $ ?

Answer 1

Hint: In the question, we are asked to evaluate an indefinite integral. The given indefinite integral is not any kind of basic or special integral which can be directly integrated, We have to employ the method of integration by parts so as to work out the integral of the given function. Integration by parts is employed generally to integrate functions that are products of two functions. But in the given question, the same method has to be applied considering $ \log (1 + {x^2}) $ as a product of $ \log (1 + {x^2}) $ and $ 1 $ .

Complete step by step solution:
In the given problem, we are required to integrate the given function, $ \log (1 + {x^2}) $ .
So, Consider $ I = \int {\log (1 + {x^2}} )dx $
We will integrate the function using integration by parts. For integration by parts, we express $ \log (1 + {x^2}) $ as product of $ \log (1 + {x^2}) $ and $ 1 $ . Hence, we get,
 $ \Rightarrow I = \int {\log (1 + {x^2}} )\left( 1 \right)dx $
Now, we apply the formula for integration by parts,
 $ \int {f\left( x \right)g\left( x \right)} dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\int {g\left( x \right)dx} } \right]dx} } $
Considering $ \log (1 + {x^2}) $ as the first function and $ \left( 1 \right) $ as the second function, we get,
 $ I = \left[ {\log (1 + {x^2}).\int {1dx} } \right] - \int {\left[ {\dfrac{{d\left( {\log (1 + {x^2})} \right)}}{{dx}}.\int {1dx} } \right]} dx $
Now, we know that integral of $ \left( 1 \right) $ with respect to x is $ \left( x \right) $ and differentiation of $ \log \left( {1 + {x^2}} \right) $ with respect to x is $ \left( {\dfrac{1}{{1 + {x^2}}}} \right)\left( {2x} \right) $ using chain rule.
\[ \Rightarrow I = \left[ {\log (1 + {x^2}).x} \right] - \int {\dfrac{{2x}}{{1 + {x^2}}}} .xdx\]
 $ \Rightarrow I = \left[ {x.\log (1 + {x^2})} \right] - 2\int {\dfrac{{{x^2}}}{{1 + {x^2}}}} dx $
 $ \Rightarrow I = \left[ {x.\log (1 + {x^2})} \right] - 2\int {\dfrac{{1 + {x^2} - 1}}{{1 + {x^2}}}} dx $
 $ \Rightarrow I = \left[ {x.\log (1 + {x^2})} \right] - 2\int {\left[ {1 - \dfrac{1}{{1 + {x^2}}}} \right]} dx $
Now, we know that integral of $ \left( 1 \right) $ with respect to x is $ \left( x \right) $ and integral of $ \left( {\dfrac{1}{{1 + {x^2}}}} \right) $ with respect to x is $ {\tan ^{ - 1}}x $ . So, we et,
 $ \Rightarrow I = \left[ {x.\log (1 + {x^2})} \right] - 2\left[ {x - {{\tan }^{ - 1}}x} \right] + C $
 $ \Rightarrow I = - 2x + 2{\tan ^{ - 1}}x + x.\log (1 + {x^2}) + C $
So, the correct answer is “ $ - 2x + 2{\tan ^{ - 1}}x + x.\log (1 + {x^2}) + C $ ”.

Note: This problem is a typical example where we can apply Integration by parts method without having any indication about the same because the given function is not a product of two functions. Care should be taken while choosing the order of the functions as it plays a crucial role in integrating the given function using integration by parts method.