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

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 ?

• I'd say partial fraction decomposition is the way to go here. When the powers get (relatively) high, integrating rational functions like these is a difficult/tedious task, regardless of the method used. Commented Mar 12 at 12:29
• But there is no other alternative other than Partial Fraction Decomposition? Commented Mar 12 at 12:46
• There are other ways as people have noted, but personally I find them all almost equally laborious. Commented Mar 12 at 13:00
• Partial Fraction Decomposition shall be a smart way to do. Use the cover-up method, not so much time is needed to get it solved. Commented Mar 12 at 13:24

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.

• Your method is absolutely correct @Vasili. But using the series expansion and integrating will be difficult for higher values of n for $(x-6)^{n}$ when $n\geq 8$. Commented Mar 12 at 12:52

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}

• I think something is wrong with the synthetic division. From your calculation, $$3x^4+5x^3+7x^2+2x+3=(x-6)(3x^3+23x^2+145x+872)+5235.$$ We can easily check it by expansion. It is not the same as $$3x^4+5x^3+7x^2+2x+3=(x-6)(3(x-6)^3+23(x-6)^2+145(x-6)+872)+5235.$$ Commented Mar 13 at 9:52
• Thank you very much! Li Kwok Keung, you are right. I haven’t finished the division. Fixed now.
– Lai
Commented Mar 13 at 10:29

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.

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}