How do you find the integral of $\int \frac{t}{t^{4} + 16}$ $int t/(t^4+16)$ ?

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How do you find the integral of $\int \frac{t}{t^{4} + 16}$ $int t/(t^4+16)$ ?

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Answer 1 · 2017-02-09T13:49:11

The answer is $= \frac{1}{8} arctan (\frac{t^{2}}{4}) + C$ $=1/8arctan(t^2/4)+C$

Explanation:

We perform this integral by substitution

Let $u = \frac{t^{2}}{4}$ $u=t^2/4$

$d u = \frac{2}{4} t d t = \frac{t}{2} d t$ $du=2/4tdt=t/2dt$

$t^{4} + 16 = 16 u^{2} + 16 = 16 (u^{2} + 1)$ $t^4+16=16u^2+16=16(u^2+1)$

Therefore,

$\int \frac{t d t}{t^{4} + 16} = \int \frac{2 d u}{16 (u^{2} + 1)}$ $int(tdt)/(t^4+16)=int(2du)/(16(u^2+1))$

$= \frac{1}{8} \int \frac{d u}{u^{2} + 1}$ $=1/8int(du)/(u^2+1)$

Let $u = tan θ$ $u=tantheta$

$d u = {sec}^{2} θ d θ$ $du=sec^2theta d theta$

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

So,

$\frac{1}{8} \int \frac{d u}{u^{2} + 1} = \frac{1}{8} \int \frac{{sec}^{2} θ d θ}{{sec}^{2} θ}$ $1/8int(du)/(u^2+1)=1/8int(sec^2 theta d theta)/(sec ^2 theta)$

$= \frac{1}{8} \int d θ = \frac{θ}{8}$ $=1/8intd theta=theta/8$

$= \frac{1}{8} arctan (u)$ $=1/8arctan(u)$

Therefore,

$\int \frac{t d t}{t^{4} + 16} = \frac{1}{8} arctan (\frac{t^{2}}{4}) + C$ $int(tdt)/(t^4+16)=1/8arctan(t^2/4)+C$

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How do you find the integral of $\int \frac{t}{t^{4} + 16}$ $int t/(t^4+16)$ ?

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