Integrate the following $\int \frac{t}{t^{4} + 2} d t$ $int t/(t^4 +2) dt$ ?

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Integrate the following $\int \frac{t}{t^{4} + 2} d t$ $int t/(t^4 +2) dt$ ?

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Answer 1 · 2017-02-19T16:35:21

$\frac{1}{2 \sqrt{2}} {tan}^{- 1} (\frac{t^{2}}{\sqrt{2}}) + C$ $1/(2sqrt2)tan^-1(t^2/sqrt2)+C$

Explanation:

Let $t^{2} = \sqrt{2} tan θ$ $t^2=sqrt2tantheta$ . This implies that $2 t d t = \sqrt{2} {sec}^{2} θ d θ$ $2tdt=sqrt2sec^2thetad theta$ . Then:

$I = \frac{1}{2} \int \frac{2 t d t}{{(t^{2})}^{2} + 2} = \frac{1}{2} \int \frac{\sqrt{2} {sec}^{2} θ d θ}{{(\sqrt{2} tan θ)}^{2} + 2}$ $I=1/2int(2tdt)/((t^2)^2+2)=1/2int(sqrt2sec^2thetad theta)/((sqrt2tantheta)^2+2)$

Continuing on and using ${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ :

$I = \frac{1}{\sqrt{2}} \int \frac{{sec}^{2} θ}{2 {tan}^{2} θ + 2} d θ = \frac{1}{2 \sqrt{2}} \int \frac{{sec}^{2} θ}{{sec}^{2} θ} d θ = \frac{1}{2 \sqrt{2}} \int d θ$ $I=1/sqrt2intsec^2theta/(2tan^2theta+2)d theta=1/(2sqrt2)intsec^2theta/sec^2thetad theta=1/(2sqrt2)intd theta$

We're working in terms of $θ$ $theta$ :

$I = \frac{1}{2 \sqrt{2}} θ + C$ $I=1/(2sqrt2)theta+C$

From $t^{2} = \sqrt{2} tan θ$ $t^2=sqrt2tantheta$ we see that $θ = {tan}^{- 1} (\frac{t^{2}}{\sqrt{2}})$ $theta=tan^-1(t^2/sqrt2)$ :

$I = \frac{1}{2 \sqrt{2}} {tan}^{- 1} (\frac{t^{2}}{\sqrt{2}}) + C$ $I=1/(2sqrt2)tan^-1(t^2/sqrt2)+C$

Answer 2 · 2017-02-19T23:20:13

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