Question

plzz help me in the problem​

Answer 1

☆ Simple Integration ☆

Question:-

• $\sf\textsf{Integrate } y = 6x+5x^2-2x^3+\dfrac{1}{x^2}$

Answer:-

• We will use one simple identity of the most basic integration:

• $\large\boxed{\int x^n\, dx = \frac{x^{n+1}}{n+1}+c}$

• We have an expression in variable x. Each term of x has a specific power. For $\frac{1}{x^2}$, we can represent it as $x^{-2}$ and use the identity.

$\displaystyle \int y\, dx \\\\\\ = \int \left( 6x+5x^2-2x^3+\frac{1}{x^2}\right)\, dx \\\\\\ = 6\int x\, dx + 5 \int x^2\, dx - 2\int x^3\, dx + \int x^{-2}\, dx \\\\\\ = 6\ \frac{x^{1+1}}{1+1}+5\ \frac{x^{2+1}}{2+1}-2\ \frac{x^{3+1}}{3+1}+\frac{x^{-2+1}}{-2+1} + c \\\\\\ 6 \times \frac{x^2}{2} + 5\times \frac{x^3}{3}-2\times \frac{x^4}{4}+\frac{x^{-1}}{-1}+c\\\\\\ = 3x^2+\frac{5}{3}x^3-\frac{1}{2}x^4-\frac{1}{x}+c$

Thus, we have the answer:-

$\Large\boxed{\sf\int y\, dx = 3x^2+\frac{5}{3}x^3-\frac{1}{2}x^4-\frac{1}{x}+c}$

Answer 2

Question:-

• $\sf\textsf{Integrate} y = 6x + 5x^2 - 2x^3 +\dfrac{1}{x^2}$

Answer:-

The identity that we can use here is given below ,

• $\sf\large{\int x^n\, dx = \frac{x^{n+1}}{n+1}+c}$

In the given question we have x as the variable. As we are aware with the fact that all numbers and variables have a certain exponent . So does x has here as $\sf\:\dfrac{1}{x^2}$

It can also be represented as

• $\sf\:x^{-2}$ .

Now by using the identity ,

• $\sf\large{\int x^n\, dx = \frac{x^{n+1}}{n+1}+c}$

We will solve this question ,

Calculations :

• $\sf\::\implies\: \int y\, dx$

• $\sf\::\implies\:\int \left( 6x+5x^2-2x^3+\frac{1}{x^2}\right)\, dx$

• $\sf\::\implies\:6\int x\, dx + 5 \int x^2\, dx - 2\int x^3\, dx + \int x^{-2}\, dx$

• $\sf\::\implies\:6\ \frac{x^{1+1}}{1+1}+5\ \frac{x^{2+1}}{2+1}-2\ \frac{x^{3+1}}{3+1}+\frac{x^{-2+1}}{-2+1} + c$

• $\sf\::\implies\:6 \times \frac{x^2}{2} + 5\times \frac{x^3}{3}-2\times \frac{x^4}{4}+\frac{x^{-1}}{-1}+c$

• $\sf\::\implies\:3x^2+\frac{5}{3}$

• $\sf\:\implies\:x^3-\frac{1}{2}x^4-\frac{1}{x}+c$

Therefore we conclude to the answer as :-

• $\sf{\pink{\int y\, dx = 3x^2+\frac{5}{3}x^3-\frac{1}{2}x^4-\frac{1}{x}+c}}$