Mathematics

$$\displaystyle \int\dfrac{t^{2}}{t^{3}+1} dt = $$


SOLUTION

Consider the following Equation.

  $$ I=\int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{{{t}^{3}}+1}}dt $$

 $$ x={{t}^{3}}+1 $$

 $$ \dfrac{dx}{dt}=2{{t}^{2}} $$

 $$ =\int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{x}*\dfrac{1}{2{{t}^{2}}}}dx $$

 $$ =\dfrac{1}{2}\ln x+C $$

 $$ =\dfrac{1}{2}\ln \left( {{t}^{3}}+1 \right)+C $$

View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
$$\displaystyle \int (sin x. sin3x. cos 2x)dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Evaluate the following definite integrals:

$$\displaystyle \int _{0}^{1/3} \dfrac {1}{\sqrt {1-x^2}}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium

lf f(a$$+$$b-x) $$=\mathrm{f}(\mathrm{x})$$ then $$\displaystyle \int_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}.\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}=$$
  • A. $$\displaystyle \frac{\mathrm{a}+\mathrm{b}}{2}\int_{\mathrm{a}}^{\mathrm{b}}$$ f(b-x)dx
  • B. $$\displaystyle \frac{\mathrm{b}-a}{2}\displaystyle \int_{\mathrm{a}}^{\mathrm{b}}$$ f(x)dx
  • C. $$(\displaystyle \mathrm{a}+\mathrm{b})\int_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}$$
  • D. $$\displaystyle \frac{\mathrm{a}+\mathrm{b}}{2}\int_{\mathrm{a}}^{\mathrm{b}}$$ f(x)dx

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate the following : $$\displaystyle\int \dfrac{1}{5-4x-3x^{2}}.dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts
$$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$$ 
where  $$\displaystyle v_{1}\left ( x \right )=\int v\left ( x \right )dx,v_{2}\left ( x \right )=\int v_{1}\left ( x \right )dx ..., v_{n}\left ( x \right )= \int v_{n-1}\left ( x \right )dx$$
Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $$\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$$ where $$\displaystyle P_{n}\left ( x \right )$$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer