Question

Evaluate $\int {\dfrac{{{x^7}}}{{{{(x + {x^4})}^2}}}dx}$.

Answer 1

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.