Question #2a34f

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Question #2a34f

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Answer 1 · 2017-08-31T20:09:46

$\int \frac{x^{2}}{9 + 16 x^{6}} d x = \frac{1}{36} arctan (\frac{4}{3} x^{3}) + constant$ $intx^2/(9+16x^6)dx = 1/36arctan(4/3x^3)+"constant"$

Explanation:

For the integral $F (x) = \int \frac{x^{2}}{9 + 16 x^{6}} d x$ $F(x)=intx^2/(9+16x^6)dx$ , sub $u = x^{3}$ $u=x^3$ and $d u = 3 x^{2} d x$ $du=3x^2dx$

Then $F (x) = \frac{1}{3} \int \frac{1}{9 + 16 u^{2}} d u = \frac{1}{27} \int \frac{1}{1 + \frac{16}{9} u^{2}} d u$ $F(x) = 1/3int1/(9+16u^2)du = 1/27int1/(1+16/9u^2)du$

For the integral $F (x)$ $F(x)$ , sub $t = \frac{4}{3} u$ $t=4/3u$ and $d t = \frac{4}{3} d u$ $dt=4/3du$

Then $F (x) = \frac{1}{36} \int \frac{1}{1 + t^{2}} d t$ $F(x) = 1/36int1/(1+t^2)dt$

$F (x)$ $F(x)$ is now in a standard form and can be evaluated as a standard integral.

$F (x) = \frac{1}{36} \int \frac{1}{1 + t^{2}} d t = \frac{1}{36} arctan t = \frac{1}{36} arctan (\frac{4}{3} u) = \frac{1}{36} arctan (\frac{4}{3} x^{3}) + constant$ $F(x)=1/36int1/(1+t^2)dt=1/36arctant = 1/36arctan(4/3u) = 1/36arctan(4/3x^3)+"constant"$

Answer 2 · 2017-09-01T16:15:45

$\frac{1}{36} {tan}^{- 1} (\frac{4}{3} x^{3}) + C$ $1/36tan^-1(4/3x^3)+C$

Explanation:

$\int \frac{x^{2} d x}{9 + 16 x^{6}}$ $int(x^2dx)/(9+16x^6)$

Use the substitution $x^{3} = \frac{3}{4} tan θ, \Rightarrow 3 x^{2} d x = \frac{3}{4} {sec}^{2} θ d θ$ $x^3=3/4tantheta,=>3x^2dx=3/4sec^2thetad theta$ .

$= \frac{1}{3} \int \frac{3 x^{2} d x}{9 + 16 {(x^{3})}^{2}} = \frac{1}{3} \int \frac{\frac{3}{4} {sec}^{2} θ d θ}{9 + 16 (\frac{9}{16} {tan}^{2} θ)}$ $=1/3int(3x^2dx)/(9+16(x^3)^2)=1/3int(3/4sec^2thetad theta)/(9+16(9/16tan^2theta))$

$= \frac{1}{4} \int \frac{{sec}^{2} θ d θ}{9 (1 + {tan}^{2} θ)}$ $=1/4int(sec^2thetad theta)/(9(1+tan^2theta))$

Recall that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ :

$= \frac{1}{36} \int d θ = \frac{1}{36} θ + C$ $=1/36intd theta=1/36theta+C$

From $x^{3} = \frac{3}{4} tan θ$ $x^3=3/4tantheta$ , note that $tan θ = \frac{4}{3} x^{3}$ $tantheta=4/3x^3$ so $θ = {tan}^{- 1} (\frac{4}{3} x^{3})$ $theta=tan^-1(4/3x^3)$ :

$= \frac{1}{36} {tan}^{- 1} (\frac{4}{3} x^{3}) + C$ $=1/36tan^-1(4/3x^3)+C$

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