Question

How do you find the integral of $\dfrac{{u - 4}}{{{u^2} + 1}}$ ?

Answer 1

Hint : Here in this question given a definite integral, we have to find the integrated value of the given function. This can be solved by substitution method means substitute denominator as t and later integrated by using the general power formula of integration. And further simplify by substituting the limit points, we get the required solution.

Complete step-by-step answer :
integration is the inverse process of differentiation. An integral which is not having any upper and lower limit is known as an indefinite integral.
Consider the given function.
$\Rightarrow \,\dfrac{{u - 4}}{{{u^2} + 1}}$
Integrate with respect to u, then
$\Rightarrow \int {\dfrac{{u - 4}}{{{u^2} + 1}}\,} du$ ---------(1)
Now separate the one fraction function into two function i.e.,
$\Rightarrow \int {\left( {\dfrac{u}{{{u^2} + 1}} - \dfrac{4}{{{u^2} + 1}}} \right)\,} du$
Integrate each term with respect to u, then
$\Rightarrow \int {\dfrac{u}{{{u^2} + 1}}\,} du - \int {\dfrac{4}{{{u^2} + 1}}\,} du$ ---------(2)
Now, consider
$\Rightarrow \int {\dfrac{u}{{{u^2} + 1}}\,} du$
Substitute or put $t = {u^2} + 1$
On differentiating t with respect to u, we get
$\dfrac{{dt}}{{du}} = 2u$
Or it can be written as
$\dfrac{{dt}}{2} = u\,du$
$\Rightarrow \int {\dfrac{{dt}}{{2t}}\,}$
Where, $\dfrac{1}{2}$ is constant then take out from the integral
$\Rightarrow \dfrac{1}{2}\int {\dfrac{{dt}}{t}\,}$
As we know, integration of $\int {\dfrac{1}{x}dx = \ln x + c}$ , then
$\Rightarrow \dfrac{1}{2}\ln \left( t \right) + {c_1}$
Substitute the t value, then
$\Rightarrow \dfrac{1}{2}\ln \left( {{u^2} + 1} \right) + {c_1}$ ----------(a)
Now, consider
$\Rightarrow \int {\dfrac{4}{{{u^2} + 1}}\,} du$
Where, 4 is constant then take out from the integral
$\Rightarrow 4\int {\dfrac{1}{{{u^2} + 1}}\,} du$
As, we know the standard formula $\int {{{\tan }^{ - 1}}x\,dx = \dfrac{1}{{1 + {x^2}}} + c}$ , then
$\Rightarrow 4{\tan ^{ - 1}}u + {c_2}$ ----------(b)
Substitute equation (a) and (b) in equation (2), then
$\Rightarrow \dfrac{1}{2}\ln \left( {{u^2} + 1} \right) + {c_1} - \,4{\tan ^{ - 1}}u + {c_2}$
Where, ${c_1}$ and ${c_2}$ are integrating constant on adding both we can write $C$ as integrating constant.
$\Rightarrow \dfrac{1}{2}\ln {\left( {{u^2} + 1} \right)_1} - \,4{\tan ^{ - 1}}u + C$
Hence, it’s a required solution.
So, the correct answer is “ $\dfrac{1}{2}\ln {\left( {{u^2} + 1} \right)_1} - \,4{\tan ^{ - 1}}u + C$ ”.

Note : By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.