How do you evaluate $\int \frac{1}{9 + x^{2}}$ $int 1/(9+x^2)$ from $[\sqrt{3}, 3]$ $[sqrt3,3]$ ?

Question

How do you evaluate $\int \frac{1}{9 + x^{2}}$ $int 1/(9+x^2)$ from $[\sqrt{3}, 3]$ $[sqrt3,3]$ ?

1 Answer

Impact of this question

1488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-11T16:52:12

$\frac{π}{36} .$ $pi/36.$

Explanation:

Let $I = \int_{\sqrt{3}}^{3} \frac{1}{x^{2} + 9} d x .$ $I=int_sqrt3^3 1/(x^2+9)dx.$

By the Fundamental Theorem of Calculus,

$\int f (x) = F (x) + C \Rightarrow \int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)$ $intf(x)=F(x)+C rArr int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)$ .

Since, $\int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} a r c tan (\frac{x}{a}) + c$ $int1/(x^2+a^2)dx=1/aarc tan(x/a)+c$ , we have,

$I = \frac{1}{3} {[a r c tan (\frac{x}{3})]}_{\sqrt{3}}^{3}$ $I=1/3[arc tan(x/3)]_sqrt3^3$

$\frac{1}{3} [a r c tan (\frac{3}{3}) - a r c tan (\frac{\sqrt{3}}{3})]$ $1/3[arc tan(3/3)-arc tan(sqrt3/3)]$

$= \frac{1}{3} [\frac{π}{4} - \frac{π}{6}]$ $=1/3[pi/4-pi/6]$

$:. I=pi/36$ .

Enjoy Maths.!

Science

Math

Humanities

... and beyond

How do you evaluate $\int \frac{1}{9 + x^{2}}$ $int 1/(9+x^2)$ from $[\sqrt{3}, 3]$ $[sqrt3,3]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate int 1/(9+x^2)∫19+x2int 1/(9+x^2) from [sqrt3,3][√3,3][sqrt3,3]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate $\int \frac{1}{9 + x^{2}}$ $int 1/(9+x^2)$ from $[\sqrt{3}, 3]$ $[sqrt3,3]$ ?