Question

How do you evaluate the integral $\int {\dfrac{{\arctan x}}{{{x^2}}}}$?

Answer 1

We are given an expression. We have to find the integral of the expression. We can use the substitution method to integrate the expression by substituting $u = \arctan x$ and $v = - \dfrac{1}{x}$. Then, apply the integration by parts method to integrate the expression. Then, again replace the value of u and v in the resultant expression.

Complete step by step solution:
We have to find the integral of $\int {\dfrac{{\arctan x}}{{{x^2}}}} dx$

First, we will apply substitution,

$\Rightarrow u = \arctan x$ ……(1)

Now, we will differentiate u with respect to x, we get:

$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\arctan x$

$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{{x^2} + 1}}$

$\Rightarrow du = \dfrac{1}{{{x^2} + 1}}dx$ ……(2)

Now, we will substitute $v$ for $- \dfrac{1}{x}$.

$\Rightarrow v = - \dfrac{1}{x}$ ……(3)

Now, we will differentiate v with respect to x, we get:

$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left( { - \dfrac{1}{x}} \right)$

$\Rightarrow \dfrac{{dv}}{{dx}} = \left( {\dfrac{1}{{{x^2}}}} \right)$

$\Rightarrow dv = \dfrac{1}{{{x^2}}}dx$ ……(4)

Now, we will apply the integration by parts method, $\int u dv = uv - \int {vdu}$

Now, substitute the values from equation (1), (2), (3) and (4), we get:

$\Rightarrow I = - \dfrac{{\arctan x}}{x} + \int {\dfrac{1}{{x\left( {{x^2} + 1} \right)}}dx}$

Now, we will apply partial fractions to simplify the expression, $\dfrac{1}{{x\left( {{x^2} + 1} \right)}}$

$\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{A}{x} + \dfrac{{Bx + C}}{{{x^2} + 1}}$

$\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{{A\left( {{x^2} + 1} \right)}}{{x\left( {{x^2} + 1} \right)}} + \dfrac{{\left( {Bx + C} \right)x}}{{x\left( {{x^2} + 1} \right)}}$

Simplify the numerator over the common denominator.

$\Rightarrow A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x = 1$

Simplify the expression, we get:

$\Rightarrow A{x^2} + A + B{x^2} + Cx = 1$

$\Rightarrow \left( {A + B} \right){x^2} + Cx + A = 1$

Now, we will equate the coefficients from each side of equation.

$\Rightarrow A + B = 0$

$\Rightarrow C = 0$

$\Rightarrow A = 1$

Now, we will substitute $A = 1$ into the expression $A + B = 0$ to compute the value of B.

$\Rightarrow 1 + B = 0$

$\Rightarrow B = - 1$

Now, substitute the values of A, B and C into partial fractions.
$\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{1}{x} + \dfrac{{ - 1x + 0}}{{{x^2} + 1}}$

$\Rightarrow \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{1}{x} - \dfrac{x}{{{x^2} + 1}}$

Now, we will substitute the above equation into the integral expression, $I = - \dfrac{{\arctan x}}{x} + \int {\dfrac{1}{{x\left( {{x^2} + 1} \right)}}dx}$

$\Rightarrow I = - \dfrac{{\arctan x}}{x} + \int {\dfrac{1}{x} - \dfrac{x}{{{x^2} + 1}}dx}$

Now, integrate the expression.

$\Rightarrow I = - \dfrac{{\arctan x}}{x} + \ln \left| x \right| - \int {\dfrac{x}{{{x^2} + 1}}dx}$

Now, we will multiply and divide the integral expression by 2.

$\Rightarrow I = - \dfrac{{\arctan x}}{x} + \ln \left| x \right| - \dfrac{1}{2}\int {\dfrac{{2x}}{{{x^2} + 1}}dx}$

Now, we will integrate the expression, by applying logarithmic rule

$\Rightarrow I = - \dfrac{{\arctan x}}{x} + \ln \left| x \right| - \dfrac{1}{2}\ln \left| {{x^2} + 1} \right| + C$

Hence the integral of $\int {\dfrac{{\arctan x}}{{{x^2}}}}$ is equal to $- \dfrac{{\arctan x}}{x} + \ln \left| x \right| - \dfrac{1}{2}\ln \left| {{x^2} + 1} \right| + C$.

Note: The students must note that we have applied the integration by parts method which is used when the integral is given as a combination of two functions. A useful rule of integration by parts is known as ILATE which is given by:
I: Inverse trigonometric functions
L: logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: exponential functions