Solveeit Logo
Login

Question

Question: The value of the integral \(\int{\dfrac{2{{x}^{3}}-1}{{{x}^{4}}+x}}dx\) is equal to and (C is consta...

The value of the integral ∫2x3−1x4+xdx\int{\dfrac{2{{x}^{3}}-1}{{{x}^{4}}+x}}dx is equal to and (C is constant)
(a)ln⁡∣x3+1x∣+C\ln \left| \dfrac{{{x}^{3}}+1}{x} \right|+C
(b)12ln⁡((x3+1)2∣x∣3)+C\dfrac{1}{2}\ln \left( \dfrac{{{\left( {{x}^{3}}+1 \right)}^{2}}}{{{\left| x \right|}^{3}}} \right)+C
(c)12ln⁡(∣x3+1∣x2)+C\dfrac{1}{2}\ln \left( \dfrac{\left| {{x}^{3}}+1 \right|}{{{x}^{2}}} \right)+C
(d) ln⁡(∣x3+1∣x2)+C\ln \left( \dfrac{\left| {{x}^{3}}+1 \right|}{{{x}^{2}}} \right)+C

Explanation

Solution

We are given an integral ∫2x3−1x4+xdx\int{\dfrac{2{{x}^{3}}-1}{{{x}^{4}}+x}}dx and we have to evaluate it which we are going to do by dividing numerator and denominator by x2{{x}^{2}} then you will find the differentiation of denominator is written in the numerator then assume the denominator as “t” and write the numerator in terms of t and then do the integration. After the integration, you will find the answer in terms of “t” so replace the value of “t” that we have assumed above and this is how we got the value of the integral.

Complete step-by-step answer :
It is given in the above problem that we have to evaluate the following integral:
∫2x3−1x4+xdx\int{\dfrac{2{{x}^{3}}-1}{{{x}^{4}}+x}}dx
Dividing the numerator and denominator by x2{{x}^{2}} we get,
∫2x3−1x2x4+xx2dx =∫2x−1x2x2+1xdx..........Eq.(1) \begin{aligned} & \int{\dfrac{\dfrac{2{{x}^{3}}-1}{{{x}^{2}}}}{\dfrac{{{x}^{4}}+x}{{{x}^{2}}}}}dx \\\ & =\int{\dfrac{2x-\dfrac{1}{{{x}^{2}}}}{{{x}^{2}}+\dfrac{1}{x}}dx}..........Eq.(1) \\\ \end{aligned}
Now, let us assume the denominator of the above integral as “t”:
x2+1x=t{{x}^{2}}+\dfrac{1}{x}=t
Taking derivative on both the sides we get,
(2x−1x2)dx=dt\left( 2x-\dfrac{1}{{{x}^{2}}} \right)dx=dt
Now, as you can see from the above equation that L.H.S is equal to the numerator of eq. (1) so we can use the above relation in eq. (1) we get,
∫dtt\int{\dfrac{dt}{t}}
We know the integration of 1t\dfrac{1}{t} with respect to t is ln⁡t\ln t.
ln⁡t+C\ln t+C
Now, substituting the value of “t” in the above expression that we have assumed above we get,
ln⁡∣x2+1x∣+C\ln \left| {{x}^{2}}+\dfrac{1}{x} \right|+C
Rearranging the above expression we get,
ln⁡∣x3+1x∣+C\ln \left| \dfrac{{{x}^{3}}+1}{x} \right|+C
Hence, the value of the given integral is equal to ln⁡∣x3+1x∣+C\ln \left| \dfrac{{{x}^{3}}+1}{x} \right|+C.
Hence, the correct option is (a).

Note : In the above solution, if you forgot to write constant C after integration then you can add it because in the options given above all have the constant C so here it won’t be a problem but in subjective question paper if you forgot to write the constant C then it will cost you marks in the exam.
One more thing, you might have thought why we put modulus in the logarithm because the domain of logarithm is always positive so modulus will vanish the negative sign of the expression written inside the logarithm.