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I have posted a link there to this thread so the OP can view my work.Integration of x/[(x)^2-4x+8] dx?

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I have posted a link there to this thread so the OP can view my work.Integration of x/[(x)^2-4x+8] dx?

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We are given to evaluate:

\(\displaystyle I=\int\frac{x}{x^2-4x+8}\,dx\)

Observing that:

\(\displaystyle \frac{d}{dx}\left(x^2-4x+8 \right)=2x-4\) we will then rewrite the integral as follows:

\(\displaystyle I=\frac{1}{2}\int\frac{2x-4+4}{x^2-4x+8}\,dx=\frac{1}{2}\int\frac{2x-4}{x^2-4x+8}\,dx+2\int\frac{1}{(x-2)^2+2^2}\,dx\)

For the first integral on the right:

Use the substitution:

\(\displaystyle u=x^2-4x+8\,\therefore\,du=(2x-4)\,du\)

For the second integral on the right:

Use the substitution:

\(\displaystyle x-2=2\tan(\theta)\,\therefore\,dx=2\sec^2(\theta)\,d\theta\)

And we obtain:

\(\displaystyle I=\frac{1}{2}\int\frac{1}{u}\,du+\int\frac{\sec^2(\theta)}{\tan^2(\theta)+1}\,d\theta\)

Applying the Pythagorean identity \(\displaystyle \tan^2(x)+1=\sec^2(x)\) on the second integral, we obtain:

\(\displaystyle I=\frac{1}{2}\int\frac{1}{u}\,du+\int\,d\theta\)

Now, using the rules of integration, we obtain:

\(\displaystyle I=\frac{1}{2}\ln|u|+\theta+C\)

Back-substituting for $u$ and $\theta$, we obtain:

\(\displaystyle I=\frac{1}{2}\ln\left|x^2-4x+8 \right|+\tan^{-1}\left(\frac{x-2}{2} \right)+C\)

Hence, we have found:

\(\displaystyle \int\frac{x}{x^2-4x+8}\,dx=\frac{1}{2}\ln\left|x^2-4x+8 \right|+\tan^{-1}\left(\frac{x-2}{2} \right)+C\)