Question

Integeration Y is equals to 6 X + 5 x square - 2 x cube + 1 upon x square

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:

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