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Question: Find \(\int {\dfrac{{2x}}{{({x^2} + 1){{({x^2} + 2)}^2}}}{\text{d}}x} \)...

Find ∫2x(x2+1)(x2+2)2dx\int {\dfrac{{2x}}{{({x^2} + 1){{({x^2} + 2)}^2}}}{\text{d}}x}

Explanation

Solution

In order to solve this question, we need to use substitution and substitute x2+2{x^2} + 2 as the new variable. Then write the integral in the terms of the new variable. Solve the integral by using the method of partial fraction. At last substitute the variable as x2+2{x^2} + 2

Complete step-by-step answer:
Let us assume that
I=∫2x(x2+1)(x2+2)2dxI = \int {\dfrac{{2x}}{{({x^2} + 1){{({x^2} + 2)}^2}}}{\text{d}}x}
Now we will substitute x2+2=t{x^2} + 2 = t and t−1=x2+1t - 1 = {x^2} + 1
dt=2xdxdt = 2xdx
Using these values, we get
I=dt(t−1)t2I = \dfrac{{dt}}{{(t - 1){t^2}}}
Now, we can solve this by partial fraction. As finding the integral in this way is very difficult we divide it into the sum of the two terms with (t−1)(t - 1) and t2{t^2} as their denominators.
Hence we can write it as
dt(t−1)t2=(At+Bt2+C(t−1))dt\dfrac{{dt}}{{(t - 1){t^2}}} = \left( {\dfrac{{At + B}}{{{t^2}}} + \dfrac{C}{{(t - 1)}}} \right)dt
Here A,B,CA,B,C are constants. Now we need to solve this equation to get A,B,CA,B,C
dt(t−1)t2=((At+B)(t−1)+Ct2t2(t−1))dt\dfrac{{dt}}{{(t - 1){t^2}}} = \left( {\dfrac{{(At + B)(t - 1) + C{t^2}}}{{{t^2}(t - 1)}}} \right)dt
=((A+C)t2+(B−A)t−Bt2(t−1))dt= \left( {\dfrac{{(A + C){t^2} + (B - A)t - B}}{{{t^2}(t - 1)}}} \right)dt
Comparing coefficients of t2,t and dt{t^2},t{\text{ and }}dt in the numerator on both sides, we get
A+C=0,B−A=0,\B=−1A + C = 0,B - A = 0,\B = -1
B=−1,A=−1,C=1B = - 1,A = - 1,C = 1
Therefore
I=∫(−t−1t2+1t−1)dtI = \int {\left( {\dfrac{{ - t - 1}}{{{t^2}}} + \dfrac{1}{{t - 1}}} \right)} dt
I=∫(1t−1−t+1t2)dtI = \int {\left( {\dfrac{1}{{t - 1}} - \dfrac{{t + 1}}{{{t^2}}}} \right)} dt
I=∫(dtt−1−(t+1)dtt2)I = \int {\left( {\dfrac{{dt}}{{t - 1}} - \dfrac{{(t + 1)dt}}{{{t^2}}}} \right)}
Now ∫dtt−1=ln⁡∣t−1∣\int {\dfrac{{dt}}{{t - 1}} = \ln \left| {t - 1} \right|}
∫tt2dt=ln⁡∣t∣\int {\dfrac{t}{{{t^2}}}dt = \ln \left| t \right|}
∫dtt2=−1t\int {\dfrac{{dt}}{{{t^2}}} = - \dfrac{1}{t}}
Hence I=ln⁡∣t−1∣+1t−ln⁡∣t∣I = \ln \left| {t - 1} \right| + \dfrac{1}{t} - \ln \left| t \right|
I=ln⁡∣t−1t∣+1t=ln⁡(x2+1x2+2)+1x2+2+cI = \ln \left| {\dfrac{{t - 1}}{t}} \right| + \dfrac{1}{t} = \ln \left( {\dfrac{{{x^2} + 1}}{{{x^2} + 2}}} \right) + \dfrac{1}{{{x^2} + 2}} + c

Note: Solving these integration questions becomes very easy if we know the basic integration and the formula methods to solve integration such as partial fraction, substitution, etc.
In case of partial fraction, we should also have the knowledge of assuming the correct numerator like we did in the question.