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$$\int \frac{ 0.287x}{x^2 - 1.456x + 1.326}dx $$

I'm stuck with rational numbers, I tried to replace $\frac 1 {1000}$.

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I have added LaTeX to your question, please let me know if I have inadvertently changed the meaning. – process91 Dec 16 '12 at 19:52
@MichaelBoratko thanks its correct I forgot the $$ thanks – Sam Dec 16 '12 at 19:55
It will be a little ugly. The quadratic at the bottom has no real roots. You will need to complete the square and make a substitution. Either that or look up in a standard source (Wikipedia?) the general formula for integrating this sort of thing. – André Nicolas Dec 16 '12 at 19:56
up vote 2 down vote accepted

Evaluation as commented by André Nicolas. $$\begin{equation*} I=\int \frac{0.287x}{x^{2}-1.456x+1.326}dx \end{equation*}$$

Complete the square in the denominator $$\begin{equation*} x^{2}-1.456x+1.326=\left( x-0.728\right) ^{2}+0.79602. \end{equation*}$$ The integral can thus be written as $$\begin{equation*} I=k\int \frac{x}{\left( x-p\right) ^{2}+q^{2}}dx, \end{equation*}$$ with $$\begin{equation*} k=0.287,p=0.728,q^{2}=0.79602. \end{equation*}$$ Make the substitution $u=x-p$ and get $$\begin{eqnarray*} I &=&k\int \frac{u+p}{u^{2}+q^{2}}du=k\left( \int \frac{u}{u^{2}+q^{2}} du+\int \frac{p}{u^{2}+q^{2}}du\right) \\ &=&k\left( \frac{1}{2}\ln \left( u^{2}+q^{2}\right) +\frac{p}{q}\arctan \frac{u}{q}\right) +C \\ &=&k\left( \frac{1}{2}\ln \left( \left( x-p\right) ^{2}+q^{2}\right) +\frac{p% }{q}\arctan \frac{x-p}{q}\right) +C. \end{eqnarray*}$$

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Alright, this one takes a while so I broke it down as best I could

Take the integral: $$\int (0.287 x)/(x^2-1.456 x+1.326) dx$$

Factor out constants: $$ = 0.287 \int {x\over(x^2-1.456 x+1.326) }dx$$

Rewrite the integrand $${x\over(x^2-1.456 x+1.326)} as {(2 x-1.456)\over(2 (x^2-1.456 x+1.326))}+{0.728\over(x^2-1.456 x+1.326)}:$$ $$= 0.287 \int{(2 x-1.456)\over(2 (x^2-1.456 x+1.326))}+{0.728\over(x^2-1.456 x+1.326))} dx $$

Integrate the sum term by term and factor out constants: $$= 0.208936 \int {1\over(x^2-1.456 x+1.326)} dx + 0.1435 \int{ (2 x-1.456)/(x^2-1.456 x+1.326)} dx $$

For the integrand $$(2 x-1.456)/(x^2-1.456 x+1.326)$$, substitute u $$ = x^2-1.456 x+1.326 and du = 2 x-1.456 dx$$: $$= 0.1435 \int {1\over u} du + 0.208936 \int {1\over(x^2-1.456 x+1.326)} dx$$

For the integrand $$1\over{(x^2-1.456 x+1.326)}$$, complete the square: $$= 0.1435 \int {1\over u }du+0.208936 \int {1\over ((x-0.728)^2+0.796016)} dx$$

For the integrand $$1\over((x-0.728)^2+0.796016)$$, substitute s $$= x-0.728 and ds = dx:$$ $$ = 0.208936 \int {1\over(s^2+0.796016) }ds + 0.1435 \int {1\over u} du $$

Factor 0.796016 from the denominator: $$= 0.208936 \int {1.25626\over(1.25626 s^2+1)} ds+ 0.1435 \int { 1\over u }du$$

Factor out constants: $$= 0.262477 \int {1\over(1.25626 s^2+1)} ds + 0.1435 \int {1\over u }du$$

For the integrand $$1/(1.25626 s^2+1)$$, substitute p = 1.12083 s and dp = 1.12083 ds: $$= 0.234181 \int {1\over(p^2+1.)} dp+ 0.1435 \int {1 \over u} du$$

The integral of $$1\over(p^2+1.)$$ is 1. tan^(-1)(1. p): = $$0.234181 tan^{(-1)}(1. p)+0.1435 \int {1\over u} du$$

The integral of 1/u is log(u): = $$0.234181 tan^{(-1)}(1. p)+0.1435 log(u)+constant$$

Substitute back for p = 1.12083 s: =$$ 0.234181 tan^{(-1)}(1.12083 s)+0.1435 log(u)+constant$$

Substitute back for s = x-0.728: =$$ 0.1435 log(u)-0.234181 tan^{(-1)}(0.815963-1.12083 x)+constant$$

Substitute back for u = x^2-1.456 x+1.326:
$$= 0.1435 log(x^2-1.456 x+1.326)-0.234181 tan^{(-1)}(0.815963-1.12083x) + constant$$

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