$\int_{0}^{ln 3} \frac{e^{3 x}}{e^{6 x} + 5} d x$ $int_0^ln3 e^(3x)/(e^(6x)+5)dx$ . Help please?

Question

$\int_{0}^{ln 3} \frac{e^{3 x}}{e^{6 x} + 5} d x$ $int_0^ln3 e^(3x)/(e^(6x)+5)dx$ . Help please?

1 Answer

Impact of this question

1265 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-02T10:23:36

$I = \frac{\sqrt{5}}{15} (arctan (\frac{27}{\sqrt{5}}) - arctan (\frac{1}{\sqrt{5}}))$ $I = sqrt(5)/15 (arctan(27/sqrt(5)) - arctan(1/(sqrt(5))))$
$I \approx 0.15915$ $I ~~ 0.15915$

Explanation:

We wish to solve $\int_{0}^{ln 3} \frac{e^{3 x}}{e^{6 x} + 5} d x$ $int_0^(ln3) (e^(3x))/(e^(6x)+5) dx$ .

We note that $e^{6 x} = {(e^{3 x})}^{2}$ $e^(6x) = (e^(3x))^2$ and make the u-substitution $u = e^{3 x}$ $u = e^(3x)$ . It follows that $d u = 3 e^{3 x} d x$ $du = 3e^(3x) dx$ .

We prepare our integral for the substitution by multiplying by $\frac{3}{3}$ $3/3$ .

$\frac{1}{3} \int_{0}^{ln 3} \frac{3 e^{3 x}}{{(e^{3 x})}^{2} + 5} d x$ $1/3 int_0^(ln3) (3e^(3x))/((e^(3x))^2 + 5) dx$

Note that our lower bound becomes $e^{3 \cdot 0} = 1$ $e^(3*0) = 1$ and our upper bound becomes $e^{3 ln 3} = e^{ln (3^{3})} = 3^{3} = 27$ $e^(3ln3) = e^(ln(3^3)) = 3^3 = 27$ .

$\frac{1}{3} \int_{1}^{27} \frac{1}{u^{2} + 5} d u$ $1/3 int_1^(27) 1 / (u^2 + 5) du$

This nearly looks like the proper form to yield $arctan (u)$ $arctan(u)$ , but the $5$ $5$ needs to be a $1$ $1$ . We factor out the $5$ $5$ and make another substitution.

$\frac{1}{3} \int_{1}^{27} \frac{1}{5 (\frac{u^{2}}{5} + 1)} d u = \frac{1}{15} \int_{1}^{27} \frac{1}{{(\frac{u}{\sqrt{5}})}^{2} + 1} d u$ $1/3 int_1^(27) 1 / (5(u^2/5 + 1))du = 1/15 int_1^(27) 1 / ((u/sqrt(5))^2 + 1) du$

Let $v = \frac{u}{\sqrt{5}}$ $v = u/sqrt(5)$ . Then $d v = \frac{1}{\sqrt{5}} d u$ $dv = 1/sqrt(5) du$ . This yields the integral:

$\frac{\sqrt{5}}{15} \int_{\frac{1}{\sqrt{5}}}^{\frac{27}{\sqrt{5}}} \frac{1}{v^{2} + 1} d v$ $sqrt(5)/15 int_(1/sqrt(5))^(27/sqrt(5)) 1 / (v^2 + 1) dv$

This is the proper form to yield $arctan (v)$ $arctan(v)$ . Evaluating yields:

$\frac{\sqrt{5}}{15} {[arctan v]}_{\frac{1}{\sqrt{5}}}^{\frac{27}{\sqrt{5}}}$ $sqrt(5)/15 [arctanv]_(1/sqrt(5))^(27/sqrt(5))$
$= \frac{\sqrt{5}}{15} (arctan (\frac{27}{\sqrt{5}}) - arctan (\frac{1}{\sqrt{5}}))$ $= sqrt(5)/15 (arctan(27/sqrt(5)) - arctan(1/sqrt(5)))$
$\approx 0.15915$ $~~ 0.15915$

Science

Math

Humanities

... and beyond

$\int_{0}^{ln 3} \frac{e^{3 x}}{e^{6 x} + 5} d x$ $int_0^ln3 e^(3x)/(e^(6x)+5)dx$ . Help please?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

∫ln30e3xe6x+5dxint_0^ln3 e^(3x)/(e^(6x)+5)dx. Help please?

1 Answer

Explanation:

Related questions

Impact of this question

$\int_{0}^{ln 3} \frac{e^{3 x}}{e^{6 x} + 5} d x$ $int_0^ln3 e^(3x)/(e^(6x)+5)dx$ . Help please?