∫4/(x^2+2x+3) dx= dan ∫x^n.ln x^n dx= dan ∫1/((x+1)(x+2)^(1/2)) dx batasnya 2 sampai 7.... Question from @hanamutiadewi

September 2018 1 70 Report

∫4/(x^2+2x+3) dx=
dan
∫x^n.ln x^n dx=
dan
∫1/((x+1)(x+2)^(1/2)) dx batasnya 2 sampai 7

Takamori37 Atas:
$\displaystyle \int \frac{4}{x^2+2x+3}\, dx=\int\frac{4}{(x+1)^2+2} \\ \text{Tips:} \\ \int\frac{1}{(x+a)^2+u^2}=\frac{1}{u}\tan^{-1}(\frac{x+a}{u}) \\ \text{Maka, dengan $u=\sqrt{2}, a=1$} \\ \int \frac{4}{x^2+2x+3}\, dx=\frac{4}{\sqrt{2}}\tan^{-1}(\frac{x+1}{\sqrt{2}})+C \\\\ \int \frac{4}{x^2+2x+3}\, dx=2\sqrt{2}\times \tan^{-1}(\frac{x+1}{\sqrt{2}})+C$

Tengah:
$\int x^n.\ln(x^n)\, dx \\ $Integral parsial:$ \\ U=x^n~~~~~~~~~~~~~~dU=n.x^{n-1}\, dx \\ dV=\ln(x^n)\, dx~~~~~V=\frac{n}{x} \\ $Sehingga:$ \\ \int x^n.\ln(x^n)\, dx=\frac{n}{x}\times x^n-\int \frac{n}{x}\times n.x^{n-1}\, dx \\ \int x^n.\ln(x^n)\, dx=n.x^{n-1}-\int n^2.x^{n-2}\, dx \\ \int x^n.\ln(x^n)\, dx=n.x^{n-1}-\frac{n^2}{n-1}x^{n-1}+C \\ \int x^n.\ln(x^n)\, dx=\frac{n(n-1)-n^2}{n-1}x^{n-1}+C \\ \int x^n.\ln(x^n)\, dx=\frac{n^2-n-n^2}{n-1}x^{n-1}+C \\ \int x^n.\ln(x^n)\, dx=\frac{n}{1-n}x^{n-1}+C$

Bawah:
$\displaystyle \int_2^7\frac{1}{((x+1)(x+2))^{1/2}}\, dx=\int_2^7\frac{1}{\sqrt{x^2+3x+2}}\, dx \\\\ \int_2^7\frac{1}{\sqrt{(x+\frac{3}{2})^2-\frac{1}{4}}}\, dx$
Dengan menggambar segitiga siku-siku:
Misalkan:
$\displaystyle \tan a=\frac{\sqrt{(x+\frac{3}{2})^2-\frac{1}{4}}}{1/2} \\\\ \sqrt{(x+\frac{3}{2})^2-\frac{1}{4}}=\frac{1}{2}\tan a \\\\\\ \sec a=\frac{x+\frac{3}{2}}{1/2} \\\\ $$x+\frac{3}{2}=\frac{1}{2}\sec a \\ \, dx=\frac{1}{2}\tan a\sec a\, da$
Maka, menjadi :
$\displaystyle \int_2^7\frac{\frac{1}{2}\tan a\sec a\, da}{\frac{1}{2}\tan a}=\int_2^7\sec a\, da=\ln|\sec a+\tan a|+C$
Substitusi ulang dari segitiga, didapat:
$\displaystyle \ln\left|\frac{x+\frac{3}{2}}{\frac{1}{2}}+\frac{\sqrt{(x+\frac{3}{2})^2-\frac{1}{4}}}{\frac{1}{2}}\right|_2^7 \\\\ \ln\left|\frac{7+\frac{3}{2}}{\frac{1}{2}}+\frac{\sqrt{(7+\frac{3}{2})^2-\frac{1}{4}}}{\frac{1}{2}}\right|-\ln\left|\frac{2+\frac{3}{2}}{\frac{1}{2}}+\frac{\sqrt{(2+\frac{3}{2})^2-\frac{1}{4}}}{\frac{1}{2}}\right|$

$\displaystyle \ln\left|\frac{\frac{17}{2}}{\frac{1}{2}}+\frac{\sqrt{(\frac{17}{2})^2-\frac{1}{4}}}{\frac{1}{2}}\right|-\ln\left|\frac{\frac{7}{2}}{\frac{1}{2}}+\frac{\sqrt{(\frac{7}{2})^2-\frac{1}{4}}}{\frac{1}{2}}\right| \\\\ \ln|17+2\times \sqrt{\frac{288}{4}}|-\ln|7+2\times \sqrt{\frac{48}{4}}| \\\\ \ln|17+2\times \sqrt{72}|-\ln|7+2\times \sqrt{12}| \\ \ln(17+12\sqrt{3})-\ln(7+4\sqrt{3}) \\ \ln\frac{17+12\sqrt{3}}{7+4\sqrt{3}}=\ln\frac{(17+12\sqrt{3})(7-4\sqrt{3})}{49-48}$

$\displaystyle \ln\frac{119-68\sqrt{3}+84\sqrt{3}-144}{1}=\ln(26\sqrt{3}-25)$