Question #0345c

Question

Question #0345c

1 Answer

Impact of this question

1320 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-13T15:08:42

If $\int \frac{(x^{2} + x + 1) d x}{x^{2} + 1} = x + \frac{1}{2} ln (x^{2} + 1) + C$ $color(red)(int ((x^2+x+1)dx)/(x^2+1)=x+1/2ln(x^2+1)+C)$

If $\int (x^{2} + x + \frac{1}{x^{2}} + 1) d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} - \frac{1}{x} + x + C$ $color(red)(int (x^2+x+1/x^2+1)dx=x^3/3+x^2/2-1/x+x+C)$

Explanation:

I can see two possibilities in the problem that you wrote.

If the problem is

$\int \frac{(x^{2} + x + 1) d x}{x^{2} + 1}$ $color(blue)(int ((x^2+x+1)dx)/(x^2+1))$

The solution is

$\int \frac{(x^{2} + x + 1) d x}{x^{2} + 1} = \int \frac{(x^{2} + 1 + x) d x}{x^{2} + 1} = \int (\frac{x^{2} + 1}{x^{2} + 1} + \frac{x}{x^{2} + 1}) d x = \int (1 + \frac{x}{x^{2} + 1}) d x$ $int ((x^2+x+1)dx)/(x^2+1)=int ((x^2+1+x)dx)/(x^2+1)=int ((x^2+1)/(x^2+1)+x/(x^2+1))dx=int (1+x/(x^2+1))dx$

$\int \frac{(x^{2} + x + 1) d x}{x^{2} + 1} = \int d x + \frac{1}{2} \int (\frac{2 x}{x^{2} + 1}) d x$ $int ((x^2+x+1)dx)/(x^2+1)=int dx+1/2int ((2x)/(x^2+1))dx$

$\int \frac{(x^{2} + x + 1) d x}{x^{2} + 1} = x + \frac{1}{2} ln (x^{2} + 1) + C$ $color(red)(int ((x^2+x+1)dx)/(x^2+1)=x+1/2ln(x^2+1)+C)$

If the problem is

$\int (x^{2} + x + \frac{1}{x^{2}} + 1) d x$ $color(blue)(int (x^2+x+1/x^2+1)dx)$

$\int (x^{2} + x + \frac{1}{x^{2}} + 1) d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} - \frac{1}{x} + x + C$ $color(red)(int (x^2+x+1/x^2+1)dx=x^3/3+x^2/2-1/x+x+C)$

God bless....I hope the explanation is useful.

Science

Math

Humanities

... and beyond

Question #0345c

1 Answer

Explanation:

Related questions

Impact of this question