Question

How do you integrate by substitution $\int {\left( {\dfrac{{{t^3}}}{3} + \dfrac{1}{{4{t^2}}}} \right)dt?}$

Answer 1

Hint : In the given question we have to do the integration of the given term. Here you can use the simple formula by which you can integrate $\left( {\dfrac{{{t^3}}}{3} + \dfrac{1}{{4{t^2}}}} \right)$ .
Formula which you can use to solve this question is:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
Just separate the term in two parts and apply the above given formula and find the answer also don’t forget to add the integral constant at the end when you will integrate the terms as it allows us to express the general form of antiderivatives.Also be careful with the integration and the negative power.

Complete step-by-step answer :
In the given question we have to find the integration of the term
$\left( {\dfrac{{{t^3}}}{3} + \dfrac{1}{{4{t^2}}}} \right)$ with respect to $t$ .
Now to find the integration you don’t have to use substitution method instead use the following formula:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
Now first of all separate the given term in two integration i.e.
$\int {\left( {\dfrac{{{t^3}}}{3}} \right)dt + \int {\dfrac{1}{4}{t^{ - 2}}} } dt$
Now integrating the above term with respect to $t$ we get:

$$\dfrac{1}{3}\dfrac{{{t^4}}}{4} + \left( {\dfrac{1}{4}} \right)\dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + c \\\ \Rightarrow \dfrac{1}{3}\dfrac{{{t^4}}}{4} + \left( {\dfrac{1}{4}} \right)\dfrac{{{t^{ - 1}}}}{{ - 1}} + c \\\ \Rightarrow \dfrac{{{t^4}}}{{12}} - \dfrac{1}{{4t}} + c \;$$

Here $c$ represents the integral constant and must be added while integrating the terms.
Hence, $\dfrac{{{t^4}}}{{12}} - \dfrac{1}{{4t}} + c$ is the integration of the term $\left( {\dfrac{{{t^3}}}{3} + \dfrac{1}{{4{t^2}}}} \right)$ with respect to $t$ .
So, the correct answer is “$\dfrac{{{t^4}}}{{12}} - \dfrac{1}{{4t}} + c$”.

Note : Here you should learn the formula used in the above question as it is a basic formula so it will be used in most of the integration questions, also instead of substitution we have used a formula to get the result. Also learn a few more basic integration formulas to solve integration questions easily and never forget to add the integral constant $c$ as it allows us to express the general form of antiderivatives.