How to integrate $\int \frac{3x^{4}+5x^{3}+7x^{2}+2x+3}{(x-6)^{5}}dx$?

Question

Q) How to Integrate $\int \frac{3x^{4}+5x^{3}+7x^{2}+2x+3}{(x-6)^{5}}dx$ ?

First of all let me tell what I think about this question.

In my Coaching Institute, the chapter 'Integration' is over. This question came in my mind while I was solving the questions of 'Integration By Partial Fraction Decomposition'.

Let me give two examples:

Example 1)

Let's integrate $\int\frac{x-5}{(x-7)^{2}}dx$

Now let me tell the solution of $\int\frac{x-5}{(x-7)^{2}}dx$

Let $I=\int\frac{x-5}{(x-7)^{2}}dx$

$\implies \frac{(x-5)}{(x-7)^{2}}=\frac{A}{(x-7)}+\frac{B}{(x-7)^{2}}$

$\implies (x-5)=Ax+(B-7A)$

Upon solving we get :

$A=1, B=2$

$\implies I=\int\frac{1}{(x-7)}dx+\int\frac{2}{(x-7)^{2}}dx$

Finally, after this step, it is easy to solve.

Now let me give the $2^{nd}$ example:

Evaluate $ I_1=\int\frac{3x^{2}+2x+4}{(x-7)^{3}}dx$

Similarly we can integrate this expression by using Partial Fraction Decomposition.

$\implies \frac{3x^{2}+2x+4}{(x-7)^{3}}=\frac{A}{(x-7)}+\frac{B}{(x-7)^{2}}+\frac{C}{(x-7)^{3}}$

$\implies (3x^{2}+2x+4)=A(x-7)^{2}+B(x-7)+C$

Upon solving we get: $A=3,B=44,C=165$

$\implies I_1=\int\frac{3}{(x-7)}dx+\int\frac{44}{(x-7)^{2}}dx+\int\frac{165}{(x-7)^{3}}dx$

After this step it is easy to integrate $I_1$.

Doubt:

But I can't understand how to integrate by using Integration By Partial Fraction Decomposition for higher powers of $x$. For e.g.:

If we want to integrate

$\int\frac{3x^{5}+8x^{4}+6x^{3}+4x^{2}+5x+4}{(x-6)^{6}}$ by using Partial Fraction Decomposition then it will be a very difficult task. Similarly

Integration of $\int \frac{3x^{4}+5x^{3}+7x^{2}+2x+3}{(x-6)^{5}}dx$ by using Partial Fraction Decomposition will be a very difficult task. It will consume huge amount of time. Is there any alternative method to integrate such types of expressions without using Partial Fraction Decomposition ?

Answer 1

Hint: in this particular case there is a shortcut. Substitute $y=x-6$. Then
$\int \frac{3x^{4}+5x^{3}+7x^{2}+2x+3}{(x-6)^{5}}dx= \\ \int \frac{3(y^4+24y^3+216y^2+864y+1296)+5(y^3+18y^2+108y+216)+7(y^2+12y+36)+2(y+6)+3x^{4}+3}{y^{5}}dy$
Divide each term by $y^5$ for easy integration.

Answer 2

Synthetic division

Performing synthetic division five times to find the remainders which are the coefficients of $(x-6)^k$ as below:

we have $$ \begin{aligned} 3 x^4+5 x^3+7 x^2+2 x+3 = 3(x-6)^4+77(x-6)^3+745(x-6)^2+3218(x-6)+5235 \end{aligned} $$ Hence $$ \begin{aligned} I= & 3 \int \frac{d x}{x-6}+77 \int \frac{d x}{(x-6)^2}+745 \int \frac{d x}{(x-6)^3} +3218 \int \frac{d x}{(x-6)^4}+5235 \int \frac{d x}{(x-6)^5} \\ = & 3 \ln |x-6|-\frac{77}{x-6}-\frac{745}{2(x-6)^2}-\frac{3218}{3(x-6)^3} -\frac{5235}{4(x-6)^4}+C \end{aligned} $$

Answer 3

Use integration by parts for answering efficiently.

To apply it effectively $\int \frac{3x^4 +5x^3 + 7x^2 + 2x+ 3}{(x-6)^5}$ take numerator as the 1st and denominator as the 2nd. So you will realise that on each integration the power of denominator will get reduced due to integration and the power of numerator will also decrease due to its continuous differentiation. So it will be reduced to a final lower power integration I hope now you can take it from here.

For example doing parts for the 1st time we obtain the following integral as $$I=\frac{3x^4 +5x^3 +7x^2 +2x + 3}{(-4)(x-6)^4} - \int \frac{12x^3 + 15x^2 +14x + 2}{(-4)(x-6)^4}$$ so repeating the process on the integral on the right side you will get a lower power and doing so the equation will get reduced into a easy integral.

Answer 4

By Differentiation

Let \begin{aligned} P(x)=3 x^4+5 x^3+7 x^2+2 x+3 = A_4(x-6)^4+A_3(x-6)^3+A_2(x-6)^2+A_1(x-6)+A_0 \end{aligned} for some constants $A_0,A_1,\cdots,A_4.$

In order to extract the coefficients of $A_k$’s, we need to differentiate the identity $k$ times at $x=6$ and obtain $$A_k=\frac {P^{(k)}(6)}{k!}$$ $$\boxed{A_0=5235, A_1=3218,A_2=745,A_3=77,A_4=3}$$ Hence we have \begin{aligned} P(x)= 3(x-6)^4+77(x-6)^3+745(x-6)^2+3218(x-6)+5235, \end{aligned} and $$ \begin{aligned} I= & 3 \int \frac{d x}{x-6}+77 \int \frac{d x}{(x-6)^2}+745 \int \frac{d x}{(x-6)^3} +3218 \int \frac{d x}{(x-6)^4}+5235 \int \frac{d x}{(x-6)^5} \\ = & 3 \ln |x-6|-\frac{77}{x-6}-\frac{745}{2(x-6)^2}-\frac{3218}{3(x-6)^3} -\frac{5235}{4(x-6)^4}+C \quad \blacksquare \end{aligned} $$