Evaluate $ \int {\dfrac{{{x^7}}}{{{{(x + {x^4})}^2}}}dx} $.
VerifiedHint: We will break the value of $ {x^7} $ in numerator. Thereafter will put $ {x^4} = t $ then differentiate with respect to $ n. $ Further, we will proceed with the integration by substitution method to arrive at the final result.
Complete step by step answer:
Let \[I = \int {\dfrac{{{x^7}}}{{{{(x + {x^4})}^2}}}} dx\]
As we know that $ {a^m} - {a^n} = {a^{m + n}} $
\[I = \int {\dfrac{{{x^4} \times {x^3}}}{{{{(1 + {x^4})}^2}}}} dx\]
Let $ 1 + {x^4} = t $
$ \Rightarrow {x^4} = t - 1 $
Differentiate with respect to $ x $
$ 4{x^3}dx = dt $
$ \Rightarrow {x^3}dx = \dfrac{{dt}}{4} $
\[ = \int {\dfrac{{t - 1}}{{{{(t)}^2}}} - \dfrac{{dt}}{4}} \]
\[ = \dfrac{1}{4}\int {\dfrac{{(t - 1)}}{{{t^2}}}} dt\]]
$ I = \dfrac{1}{4}\int {\left( {\dfrac{t}{{{t^2}}} - \dfrac{1}{{{t^2}}}} \right)} dt $
$ = \dfrac{1}{4}\left( {\dfrac{1}{t} - \dfrac{1}{{{t^2}}}} \right)dt $
$ I = \dfrac{1}{4}\int {\dfrac{1}{t}dt - \dfrac{1}{4}\int {\dfrac{1}{{{t^2}}}dt} } $
\[\dfrac{1}{4}\int {\dfrac{1}{t}dt - \dfrac{1}{4}\int {{t^{ - 2}}dt} } \]
As we know that \[\int {\dfrac{1}{x}dx = \log x} \]
Then,
$ I = \dfrac{1}{4}\log t - \dfrac{1}{4}\left( {\dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) $
\[I = \dfrac{1}{4}\log t - \dfrac{1}{4} \times \left( {\dfrac{{{t^{ - 1}}}}{{ - 1}}} \right)\]
$ = \dfrac{1}{4}\log t + \dfrac{1}{x}{t^{ - 1}} $ I
$ I = \dfrac{1}{4}\log t + \dfrac{1}{4} \times \dfrac{1}{t} $
Again, we will substitute the $ t $ in the form of $ x $ .
$ I = \dfrac{1}{4}\log (1 + {x^4}) + \dfrac{1}{4} \times \left( {\dfrac{1}{{1 + {x^4}}}} \right) + C $
$ I = \dfrac{1}{4}\left[ {\log (1 + {x^4}) + \dfrac{1}{{(1 + {x^4})}}} \right] + C $
Note: Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. Students should know that \[\int {\dfrac{1}{x}dx = \log x} \] and \[\int {{x^x}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]and put the value of $ t $ carefully otherwise, we will get the wrong answer.
