
Find $\int {\dfrac{{2x}}{{({x^2} + 1){{({x^2} + 2)}^2}}}{\text{d}}x} $
Answer
588.9k+ views
Hint: In order to solve this question, we need to use substitution and substitute ${x^2} + 2$ 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 ${x^2} + 2$
Complete step-by-step answer:
Let us assume that
$I = \int {\dfrac{{2x}}{{({x^2} + 1){{({x^2} + 2)}^2}}}{\text{d}}x} $
Now we will substitute ${x^2} + 2 = t$ and $t - 1 = {x^2} + 1$
$dt = 2xdx$
Using these values, we get
$I = \dfrac{{dt}}{{(t - 1){t^2}}}$
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 $(t - 1)$ and ${t^2}$ as their denominators.
Hence we can write it as
$\dfrac{{dt}}{{(t - 1){t^2}}} = \left( {\dfrac{{At + B}}{{{t^2}}} + \dfrac{C}{{(t - 1)}}} \right)dt$
Here $A,B,C$ are constants. Now we need to solve this equation to get $A,B,C$
$\dfrac{{dt}}{{(t - 1){t^2}}} = \left( {\dfrac{{(At + B)(t - 1) + C{t^2}}}{{{t^2}(t - 1)}}} \right)dt$
$ = \left( {\dfrac{{(A + C){t^2} + (B - A)t - B}}{{{t^2}(t - 1)}}} \right)dt$
Comparing coefficients of ${t^2},t{\text{ and }}dt$ in the numerator on both sides, we get
$A + C = 0,B - A = 0,\B = -1$
$B = - 1,A = - 1,C = 1$
Therefore
$I = \int {\left( {\dfrac{{ - t - 1}}{{{t^2}}} + \dfrac{1}{{t - 1}}} \right)} dt$
$I = \int {\left( {\dfrac{1}{{t - 1}} - \dfrac{{t + 1}}{{{t^2}}}} \right)} dt$
$I = \int {\left( {\dfrac{{dt}}{{t - 1}} - \dfrac{{(t + 1)dt}}{{{t^2}}}} \right)} $
Now $\int {\dfrac{{dt}}{{t - 1}} = \ln \left| {t - 1} \right|} $
$\int {\dfrac{t}{{{t^2}}}dt = \ln \left| t \right|} $
$\int {\dfrac{{dt}}{{{t^2}}} = - \dfrac{1}{t}} $
Hence $I = \ln \left| {t - 1} \right| + \dfrac{1}{t} - \ln \left| t \right|$
$I = \ln \left| {\dfrac{{t - 1}}{t}} \right| + \dfrac{1}{t} = \ln \left( {\dfrac{{{x^2} + 1}}{{{x^2} + 2}}} \right) + \dfrac{1}{{{x^2} + 2}} + c$
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.
Complete step-by-step answer:
Let us assume that
$I = \int {\dfrac{{2x}}{{({x^2} + 1){{({x^2} + 2)}^2}}}{\text{d}}x} $
Now we will substitute ${x^2} + 2 = t$ and $t - 1 = {x^2} + 1$
$dt = 2xdx$
Using these values, we get
$I = \dfrac{{dt}}{{(t - 1){t^2}}}$
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 $(t - 1)$ and ${t^2}$ as their denominators.
Hence we can write it as
$\dfrac{{dt}}{{(t - 1){t^2}}} = \left( {\dfrac{{At + B}}{{{t^2}}} + \dfrac{C}{{(t - 1)}}} \right)dt$
Here $A,B,C$ are constants. Now we need to solve this equation to get $A,B,C$
$\dfrac{{dt}}{{(t - 1){t^2}}} = \left( {\dfrac{{(At + B)(t - 1) + C{t^2}}}{{{t^2}(t - 1)}}} \right)dt$
$ = \left( {\dfrac{{(A + C){t^2} + (B - A)t - B}}{{{t^2}(t - 1)}}} \right)dt$
Comparing coefficients of ${t^2},t{\text{ and }}dt$ in the numerator on both sides, we get
$A + C = 0,B - A = 0,\B = -1$
$B = - 1,A = - 1,C = 1$
Therefore
$I = \int {\left( {\dfrac{{ - t - 1}}{{{t^2}}} + \dfrac{1}{{t - 1}}} \right)} dt$
$I = \int {\left( {\dfrac{1}{{t - 1}} - \dfrac{{t + 1}}{{{t^2}}}} \right)} dt$
$I = \int {\left( {\dfrac{{dt}}{{t - 1}} - \dfrac{{(t + 1)dt}}{{{t^2}}}} \right)} $
Now $\int {\dfrac{{dt}}{{t - 1}} = \ln \left| {t - 1} \right|} $
$\int {\dfrac{t}{{{t^2}}}dt = \ln \left| t \right|} $
$\int {\dfrac{{dt}}{{{t^2}}} = - \dfrac{1}{t}} $
Hence $I = \ln \left| {t - 1} \right| + \dfrac{1}{t} - \ln \left| t \right|$
$I = \ln \left| {\dfrac{{t - 1}}{t}} \right| + \dfrac{1}{t} = \ln \left( {\dfrac{{{x^2} + 1}}{{{x^2} + 2}}} \right) + \dfrac{1}{{{x^2} + 2}} + c$
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.
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