Question

Find the value of $\int{\dfrac{3x+1}{{{\left( 3{{x}^{2}}+2x+1 \right)}^{3}}}dx}$.

Answer 1

We must first assume the integral to be equal to a variable (say $I$) and then we can use the substitution method to solve this integration. We must substitute the denominator term and then we can easily solve this problem. We just use the integration formula $\int{{{y}^{m}}dy=\dfrac{{{y}^{m+1}}}{m+1}}+c$.

Complete step-by-step solution:
We need to find the value $I=\int{\dfrac{3x+1}{{{\left( 3{{x}^{2}}+2x+1 \right)}^{3}}}dx}$.
Let us try to solve this question using the concept of integration by substitution. Let us assume a variable t that is equal to the denominator in this question.
Hence, we have, $t=3{{x}^{2}}+2x+1$.
Let us now try to differentiate the variable t. Thus, we get
$dt=d\left( 3{{x}^{2}}+2x+1 \right)$
We can also write the above equation as
$dt=d\left( 3{{x}^{2}} \right)+d\left( 2x \right)+d\left( 1 \right)$
On differentiating these terms individually, we get
$dt=6xdx+2dx+0$
On taking 2 as common from the right hand side, we get
$dt=2\left( 3x+1 \right)dx$
So, we can also write the above equation as
$\left( 3x+1 \right)dx=\dfrac{1}{2}dt$
Thus, the required becomes,
$I=\int{\dfrac{1}{{{t}^{3}}}\times \dfrac{1}{2}dt}$
Or, we can also write this as
$I=\dfrac{1}{2}\int{\dfrac{1}{{{t}^{3}}}dt}$
Also, we know that, $\dfrac{1}{{{a}^{m}}}={{a}^{-m}}$.
Thus, the integral becomes
$I=\dfrac{1}{2}\int{{{t}^{-3}}dt}$
We all know very well that the formula for integration of $y$ raised to the power m is given as $\int{{{y}^{m}}dy=\dfrac{{{y}^{m+1}}}{m+1}}+c$
Using this formula, we can easily integrate our required integral. Thus, we get
$I=\dfrac{1}{2}\left( \dfrac{{{t}^{-3+1}}}{-3+1} \right)+k$, where k is the constant of integration.
On simplifying, we get
$I=\dfrac{1}{2}\left( \dfrac{{{t}^{-2}}}{-2} \right)+k$
We can also write the above equation as
$I=-\dfrac{1}{4{{t}^{2}}}+k$
We can now substitute the value of t back into this equation. Thus, we get
$I=-\dfrac{1}{4{{\left( 3{{x}^{2}}+2x+1 \right)}^{2}}}+k$.
Hence, the value of $\int{\dfrac{3x+1}{{{\left( 3{{x}^{2}}+2x+1 \right)}^{3}}}dx}$ is equal to $-\dfrac{1}{4{{\left( 3{{x}^{2}}+2x+1 \right)}^{2}}}+k$.

Note: We must take care so as not to forget the constant of integration in our final answer. We must understand that it is always advisable to assume the required integral to any variable, like $I$ in this case. This will help us not to neglect the terms outside the integration, like $\dfrac{1}{2}$ in this case.