How to integrate $$\int_{2}^4 \frac{1+x}{1+2x^{2}+x^{4}}dx $$ I am trying to integrate the above expression by writing the expression as $$\int_{2}^4 \frac{1}{1+2x^{2}+x^{4}}dx +\int_{2}^4 \frac{x}{1+2x^{2}+x^{4}}dx $$ Now I can easily integrate $$\int_{2}^4 \frac{x}{1+2x^{2}+x^{4}}dx $$ To integrate this we have to substitute $x^{2}$ $=$ $u$. Now $x×dx$ $=$ $du/2$ Because, $1+2x^{2}+x^{4}$ can be written as $(x^{2}+1)^{2}$. But I am facing problem to integrate $$\int_{2}^4 \frac{1}{1+2x^{2}+x^{4}}dx $$ Please help me out to integrate.
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Notice the denominator is $(x^2+1)^2=(x+i)^2(x-i)^2$. Now substitute $x=\tan(\theta)$ or use partial fractions. – Тyma Gaidash Oct 30 '23 at 17:02
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Integrate, without breaking up the integral, as follows \begin{align} &\int_{2}^4 \frac{1+x}{1+2x^{2}+x^{4}}dx\\ =& \int_{2}^4 \frac1{2(x-1)}d\bigg[ \frac{(x-1)^2}{1+x^2}\bigg]\\ \overset{ibp}=&-\frac1{85}+\frac12\int_2^4\frac1{1+x^2}dx =-\frac1{85}+\frac12\tan^{-1}\frac29 \end{align}
Quanto
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\begin{align} &\int_{2}^4 \frac{1+x}{1+2x^{2}+x^{4}}dx\\ \overset{x=\tan \theta}=& \int_{\arctan 2}^{\arctan 4} \frac{1+\tan\theta}{\sec^4\theta}\sec^2\theta d\theta\\ =&\int_{\arctan 2}^{\arctan 4} (\cos^2\theta+\sin\theta\cos\theta)d\theta\\ =&\frac{\theta}2-\frac12\cos^2(t)+\frac14\sin(\theta)\bigg|_{\arctan 2}^{\arctan 4}\\ =&-\frac1{85}+\frac12(\arctan4-\arctan2). \end{align}
xpaul
- 44,000
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$$ \begin{aligned}\int_2^4 \frac{1+x}{1+2 x^2+x^4} d x = & \int_2^4 \frac{1+x}{\left(x^2+1\right)^2} d x \\ = & \frac{1}{2} \int_2^4\left(1+\frac{1}{x}\right) d\left(1-\frac{1}{x^2+1}\right) \\ = & \frac{1}{2}\left[\left(1+\frac{1}{x}\right)\left(\frac{x^2}{x^2+1}\right)\right]_2^4+\frac{1}{2} \int_2^4 \frac{1}{x^2+1} d x \\ = & -\frac{1}{85}+\frac{1}{2}\left(\tan ^{-1} 4-\tan ^{-1} 2\right) \end{aligned} $$
Lai
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