Question

How do you integrate $\dfrac{10}{(x-1)({{x}^{2}}+9)}dx$ ?

Answer 1

There are different methods to solve the integral. This question on integration will be solved by the method of sum rule. The function being a big one it is integrated one by one. While performing sum rule, at every step we need to equate L.H.S and R.H.S. We need to find the LCM of the fractions to solve the problem.

Complete step-by-step solution:
The first step to start solving is :
$\dfrac{10}{(x-1)({{x}^{2}}+9)}=\dfrac{A}{(x-1)}+\dfrac{Bx+C}{({{x}^{2}}+9)}$
We are supposed to find the value of $A,B,C$ by equating the number.
$\dfrac{10}{(x-1)({{x}^{2}}+9)}=\dfrac{A({{x}^{2}}+9)}{(x-1)}+\dfrac{\left( Bx+C \right)(x-1)}{({{x}^{2}}+9)}$
We will equate the numerator, since the denominator both the side are same:
$10=A{{x}^{2}}+9A+B{{x}^{2}}-Bx+Cx-C$
$\Rightarrow 10={{x}^{2}}(A+B)+(C-B)x+9A-C$
On equating the R.H.S with L.H.S, we get:
Since there is no ${{x}^{2}}$ term in L.H.S, the below step is performed:
$(A+B)=0$
Since there is no $x$ term in L.H.S, the below step is performed:
$C-B=0$
Constant is present in L.H.S, the below step is performed:
$9A-C=10$
On solving the three equation, we get
$A=-B$
$C=B$
Therefore
$A=-C$
So, on applying the same condition in the third equation,
$9A-C=10$
$\Rightarrow 9A-(-A)=10$
$\Rightarrow 10A=10$
$\Rightarrow A=\dfrac{10}{10}$
$\Rightarrow A=1$
So the value of $A$ to be $1$, so other terms are $B=-1$ as $(B=-A)$ and $C=-1$ as $(C=B)$ .
So the function which is to be integrated can be written as:
$\dfrac{10}{(x-1)({{x}^{2}}+9)}=\dfrac{1}{(x-1)}+\dfrac{-1x-1}{({{x}^{2}}+9)}$
$\Rightarrow \dfrac{10}{(x-1)({{x}^{2}}+9)}=\dfrac{1}{(x-1)}+\dfrac{-1-x}{({{x}^{2}}+9)}$
On integrating the function, we get:
$\Rightarrow \int{\dfrac{10}{(x-1)({{x}^{2}}+9)}dx}=\int{\dfrac{1}{(x-1)}}dx-\int{\dfrac{1x+1}{({{x}^{2}}+9)}}dx$
On expanding:
$\Rightarrow \int{\dfrac{10}{(x-1)({{x}^{2}}+9)}dx}=\int{\dfrac{1}{(x-1)}}dx-\int{\dfrac{x}{({{x}^{2}}+9)}}dx-\int{\dfrac{1}{({{x}^{2}}+9)}dx}$
Solving the equation one by one, firstly solving $\int{\dfrac{1}{(x-1)}}dx$
Integration of $\int{\dfrac{1}{x}}$ is $\ln x$. Applying the same formula in the above function we get :
$\int{\dfrac{1}{(x-1)}}dx=\ln (x-1)$ …………………………………………………………………………(i)
Next, on solving $\int{\dfrac{x}{({{x}^{2}}+9)}}$, we get:
For integration of $\dfrac{x}{({{x}^{2}}+9)}$, we need to consider a variable $t$
$t={{x}^{2}}+9$
On differentiating $t$ with respect to $x$ ,we get
$\Rightarrow \dfrac{dt}{dx}=\dfrac{d({{x}^{2}}+9)}{dx}$
$\Rightarrow \dfrac{dt}{dx}=2x$
$\Rightarrow \dfrac{1}{2}dt=xdx$
Substituting the value of $xdx$ as $\dfrac{1}{2}dt$, we get :
$\Rightarrow \int{\dfrac{dt}{2t}}=\dfrac{1}{2}\int{\dfrac{dt}{t}}$
Applying the formula we get,
$\Rightarrow \dfrac{1}{2}\ln t$
Putting the value of $t$ as $({{x}^{2}}+9)$,we get
$\int{\dfrac{x}{({{x}^{2}}+9)}}dx=\dfrac{1}{2}\ln ({{x}^{2}}+9)$ ……………………………………….(ii)
Now integrating the function $\dfrac{1}{({{x}^{2}}+9)}$, were get
$\Rightarrow \int{\dfrac{1}{({{x}^{2}}+9)}}$
$\Rightarrow \int{\dfrac{1}{3{{\left( \dfrac{x}{3} \right)}^{2}}+1}}$
As per the formula this function on integrating becomes
$\int{\dfrac{1}{({{x}^{2}}+9)}}=\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{x}{3} \right)$…………………….(iii)
Equating all the three (i),(ii),(iii) equation
$\int{\dfrac{10}{(x-1)({{x}^{2}}+9)}dx}=\int{\dfrac{1}{(x-1)}}dx-\int{\dfrac{x}{({{x}^{2}}+9)}}dx-\int{\dfrac{1}{({{x}^{2}}+9)}dx}$
$\int{\dfrac{10}{(x-1)({{x}^{2}}+9)}dx}=\ln (x-1)-\dfrac{1}{2}\ln ({{x}^{2}}+9)-\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{x}{3} \right)+c$
$\therefore \int{\dfrac{10}{(x-1)({{x}^{2}}+9)}dx}$ is $\ln (x-1)-\dfrac{1}{2}\ln ({{x}^{2}}+9)-\dfrac{1}{3}{{\tan }^{-1}}\left( \dfrac{x}{3} \right)+c$

Note: For solving questions like this all the formulas should be known to you. There are a lot of methods to integrate the function. Each function is integrated with a different method. With practice we are able to find the method for the integration of certain questions.