$\int \frac{ 0.287x}{x^2 - 1.456x + 1.326}dx $
I'm stuck with rational numbers, I tried to replace $\frac 1 {1000}$.
$\int \frac{ 0.287x}{x^2 - 1.456x + 1.326}dx $
I'm stuck with rational numbers, I tried to replace $\frac 1 {1000}$.
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 $\begin{eqnarray*} I &=&k\int \frac{u+p}{u^{2}+q^{2}}du=k\left( \int \frac{u}{u^{2}+q^{2}} du+\int \frac{p}{u^{2}+q^{2}}du\right) \\ &=&k\left( \frac{1}{2}\ln \left( u^{2}+q^{2}\right) +\frac{p}{q}\arctan \frac{u}{q}\right) +C \\ &=&k\left( \frac{1}{2}\ln \left( \left( x-p\right) ^{2}+q^{2}\right) +\frac{p% }{q}\arctan \frac{x-p}{q}\right) +C. \end{eqnarray*}$
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:
$= 0.1435 log(x^2-1.456 x+1.326)-0.234181 tan^{(-1)}(0.815963-1.12083x) + constant$