Question

Find the solution of the integral $\int {\dfrac{{{x^3} - 1}}{{{x^3} + x}}} dx$.
A. $x - \log x + \log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + c$
B. $x - \log x + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + c$
C. $x + \log x + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + {\tan ^{ - 1}}x + c$
D. $x + \log x - \dfrac{1}{2}\log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + c$

Answer 1

Here, in the given question, we have to integrate $\dfrac{{{x^3} - 1}}{{{x^3} + x}}$ with respect to $'x'$. Let us assume the value of $\int {\dfrac{{{x^3} - 1}}{{{x^3} + x}}} dx$ is equal to $I$. We will first simplify the integrand and by using partial fractions we will proceed, then we will integrate the functions individually using integration formulas.

Complete step by step answer:
Let $I = \int {\dfrac{{{x^3} - 1}}{{{x^3} + x}}} dx$
Add and subtract $x$ in the numerator.
$\Rightarrow I = \int {\dfrac{{{x^3} + x - x - 1}}{{{x^3} + x}}} dx$
We can write it as,
$\Rightarrow I = \int {1 + \dfrac{{\left( { - x - 1} \right)}}{{{x^3} + x}}} dx$
On taking negative sign as common, we get
$\Rightarrow I = \int {1 - \dfrac{{\left( {x + 1} \right)}}{{{x^3} + x}}} dx$

On splitting the integral, we get
$\Rightarrow I = \int {1dx - \int {\dfrac{{x + 1}}{{{x^3} + x}}} dx}$
$\Rightarrow I = \int {1dx - \int {\dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}} dx} .....\left( i \right)$
Let us first solve $\dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}$.
To solve this, we need to break the function into partial fractions. The way to do this is to find constants $A$, $B$ and $C$ such that
$\Rightarrow \dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}} = \dfrac{A}{x} + \dfrac{{Bx + C}}{{{x^2} + 1}}$
Take LCM on the right-hand side
$\Rightarrow \dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}} = \dfrac{{A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x}}{{x\left( {{x^2} + 1} \right)}}$

On canceling out common terms, we get
$\Rightarrow x + 1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x$
This yields $A{x^2} + A + B{x^2} + Cx = x + 1$; Then $A + B = 0$, $C = 1$ and $A = 1$, which gives $B = - 1$.
Therefore, our integral $\dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}} = \dfrac{A}{x} + \dfrac{{Bx + C}}{{{x^2} + 1}}$, becomes:
$\Rightarrow \int {\dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}dx} = \int {\left( {\dfrac{1}{x} + \dfrac{{ - x + 1}}{{{x^2} + 1}}} \right)} dx$
From $\left( i \right)$, we get
$I = \int {1dx - \int {\dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}} dx} .....\left( i \right)$

On substituting value of $\int {\dfrac{{x + 1}}{{x\left( {{x^2} + 1} \right)}}dx} = \int {\left( {\dfrac{1}{x} + \dfrac{{ - x + 1}}{{{x^2} + 1}}} \right)} dx$, we get
$\Rightarrow I = \int {1dx - \int {\left( {\dfrac{1}{x} + \dfrac{{ - x + 1}}{{{x^2} + 1}}} \right)dx} }$
$\Rightarrow I = \int {1dx - \int {\left( {\dfrac{1}{x} - \dfrac{x}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 1}}} \right)dx} }$
On opening the bracket and changing the signs (as there is negative sign outside the bracket), we get
$\Rightarrow I = \int {1dx - \int {\dfrac{1}{x}} dx + \int {\dfrac{x}{{{x^2} + 1}}dx} - \int {\dfrac{1}{{{x^2} + 1}}dx} }$
Multiply and divide $\int {\dfrac{x}{{{x^2} + 1}}dx}$, by $2$.
$\Rightarrow I = \int {1dx - \int {\dfrac{1}{x}} } dx + \dfrac{1}{2}\int {\dfrac{{2x}}{{{x^2} + 1}}} dx - \int {\dfrac{1}{{{x^2} + 1}}} dx....\left( {ii} \right)$

Now, we will integrate all the functions one-by-one.
As we know, the integral of $1$ is $x$ and the integral of $\dfrac{1}{x}$ is $\log x$.
Now let us find the integral of $\dfrac{{2x}}{{{x^2} + 1}}$.
Let ${x^2} + 1 = u$.
$\Rightarrow 2xdx = du$, from here we get
$\Rightarrow \int {\dfrac{{2x}}{{{x^2} + 1}}dx = \int {\dfrac{{du}}{u}} }$
$\Rightarrow \int {\dfrac{{du}}{u}} = \log u + {c_1}$
On substituting value of $u$, we get
$\Rightarrow \int {\dfrac{{2x}}{{{x^2} + 1}}} = \log \left( {{x^2} + 1} \right) + {c_1}$

Now, let us find the integral of $\dfrac{1}{{{x^2} + 1}}$. As we know $\int {\dfrac{1}{{{a^2} + {x^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$
$\int {\dfrac{1}{{{x^2} + 1}}} dx = {\tan ^{ - 1}}x + {c_2}$
On substituting all the values of integrals in $\left( {ii} \right)$, we get
$\Rightarrow I = \int {1dx - \int {\dfrac{1}{x}} } dx + \dfrac{1}{2}\int {\dfrac{{2x}}{{{x^2} + 1}}} dx - \int {\dfrac{1}{{{x^2} + 1}}} dx....\left( {ii} \right)$
$\Rightarrow I = x - \log x + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + {c_1} + {c_2}$
We can write ${c_1}$ and ${c_2}$ as common constant $c$.
$\therefore I = x - \log x + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) - {\tan ^{ - 1}}x + c$
Where $c$ is the arbitrary constant called constant of integration.

Therefore, the correct option is B.

Note: We have indefinite integral that is why we added integration constant. If we have a definite integral we do not add integration constant. We know integration formulas which yield inverse trigonometric functions:
$\Rightarrow \int {\dfrac{1}{{{a^2} + {x^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$
$\Rightarrow \int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$
$\Rightarrow \int {\dfrac{1}{{x\sqrt {{x^2} - {a^2}} }}} = \dfrac{1}{a}{\sec ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C$