Question

The $\int{\dfrac{1}{x+{{x}^{5}}}dx}=f\left( x \right)+c$, then the value of $\int{\dfrac{{{x}^{4}}}{x+{{x}^{5}}}dx}$ is
A. $\log x-f\left( x \right)+c$
B. $f\left( x \right)+\log x+c$
C. $f\left( x \right)-\log x+c$
D. None of these

Answer 1

Hint: We first explain the terms $\dfrac{dy}{dx}$ where $y=f\left( x \right)$.We break the given expression and then need to integrate the equation once to find all the solutions of the integration. We take one arbitrary constant term for the integration.

Complete step by step answer:
We need to find the integral of $\int{\dfrac{{{x}^{4}}}{x+{{x}^{5}}}dx}$. We have
$\dfrac{{{x}^{4}}}{x+{{x}^{5}}} =\dfrac{1+{{x}^{4}}-1}{x\left( 1+{{x}^{4}} \right)} \\
\Rightarrow \dfrac{{{x}^{4}}}{x+{{x}^{5}}}=\dfrac{1+{{x}^{4}}}{x\left( 1+{{x}^{4}} \right)}-\dfrac{1}{x+{{x}^{5}}} \\
\Rightarrow \dfrac{{{x}^{4}}}{x+{{x}^{5}}}=\dfrac{1}{x}-\dfrac{1}{x+{{x}^{5}}} \\ $
We can form $\int{\dfrac{{{x}^{4}}}{x+{{x}^{5}}}dx}$ as
\[\int{\left( \dfrac{1}{x}-\dfrac{1}{x+{{x}^{5}}} \right)dx}=\int{\dfrac{dx}{x}}-\int{\dfrac{dx}{x+{{x}^{5}}}}\]
Now we use the integral theorem of \[\int{\dfrac{dx}{x}}=\log \left| x \right|+c\]. Given $\int{\dfrac{1}{x+{{x}^{5}}}dx}=f\left( x \right)+c$.
\[\int{\dfrac{{{x}^{4}}}{x+{{x}^{5}}}dx} =\int{\dfrac{dx}{x}}-\int{\dfrac{dx}{x+{{x}^{5}}}} \\
\therefore \int{\dfrac{{{x}^{4}}}{x+{{x}^{5}}}dx}=\log \left| x \right|-f\left( x \right)+c \]

Hence, the correct option is A.

Note: The breaking of the function is necessary for the integration as chain rule can not be applied separately. If the particular value for the variable $x$ is not available then we have to use the modulus form for the integration.