oblicz całki

[2001]

a) ∫ \(\sqrt{x^2-6x+5}\) dx

b) ∫ \( \frac{1}{x} \sqrt{ \frac{1- \sqrt[3]{x} }{1+ \sqrt[3]{x} } } \)

Proszę o pełne rozwiązanie

[korki_fizyka]

a) \(x^2-6x+5 = (x -1)(x -5)\)
podstawienie: \(\sqrt{x^2-6x+5} = t(x-1)\)
b) spróbuj \(x = t^6\)

[Robakks]

a)
Mniej obliczeń będzie jeśli użyje podstawienia
\(\sqrt{x^2-6x+5}=t-x\)
\(
x^2-6x+5=t^2-2tx+x^2\\
-6x+5=t^2-2tx\\
2tx-6x=t^2-5\\
x\left(2t-6\right)=t^2-5\\
x=\frac{t^2-5}{2t-6}\\
\mbox{d}x=\frac{2t\left(2t-6\right)-2\left(t^2-5\right)}{\left(2t-6\right)^2}\mbox{d}t\\
\mbox{d}x=\frac{2t^2-12t+10}{\left(2t-6\right)^2}\mbox{d}t\\
t-x=t-\frac{t^2-5}{2t-6}\\
t-x=\frac{t^2-6t+5}{2t-6}\\
2\int{\frac{\left(t^2-6t+5\right)^2}{\left(2t-6\right)^3}\mbox{d}t}
\)

b)

Ten pierwiastek wygląda jak wzór na tangens połowy kąta więc można podstawić

\(x=\cos^{3}{\left(t\right)}\\
\mbox{d}x=-3\cos^{2}{\left(t\right)}\sin{\left(t\right)}\mbox{d}t\\
-3\int{\frac{1}{\cos^{3}{\left(t\right)}}\cdot\tan{\left(\frac{t}{2}\right)}\cdot\cos^{2}{\left(t\right)}\sin{\left(t\right)}\mbox{d}t}\\
-3\int{\tan{\left(t\right)}\cdot\tan{\left(\frac{t}{2}\right)}\mbox{d}t}\\
-6\int{\frac{\tan^{2}{\left(\frac{t}{2}\right)}}{1-\tan^{2}{\left(\frac{t}{2}\right)}}\mbox{d}t}\\
u=\tan{\left(\frac{t}{2}\right)}\\
\frac{t}{2}=\arctan{u}\\
t=2\arctan{u}\\
\mbox{d}t=\frac{2}{1+u^2}\mbox{d}u\\
-12\int{\frac{u^2}{\left(1-u^2\right)\left(1+u^2\right)}\mbox{d}u}\\
-6\int{\frac{\left(1+u^2\right)-\left(1-u^2\right)}{\left(1-u^2\right)\left(1+u^2\right)}\mbox{d}u}\\
-6\int{\frac{1}{1-u^2}\mbox{d}u}+6\int{\frac{1}{1+u^2}\mbox{d}u}\\
-3\int{\frac{\left(1-u\right)+\left(1+u\right)}{\left(1-u\right)\left(1+u\right)}\mbox{d}u}+6\int{\frac{1}{1+u^2}\mbox{d}u}\\
-3\int{\frac{1}{1+u}\mbox{d}u}-3\int{\frac{1}{1-u}\mbox{d}u}+6\int{\frac{1}{1+u^2}\mbox{d}u}\\
3\ln{\left|\frac{1-u}{1+u}\right|}+6\arctan{\left(u\right)}+C_{1}\\
3\ln{\left|\frac{\left(1-u\right)^2}{\left(1+u\right)\left(1-u\right)}\right|}+6\arctan{\left(u\right)}+C_{1}\\
3\ln{\left|\frac{\left(1+u^2-2u\right)}{1-u^2}\right|}+6\arctan{\left(u\right)}+C_{1}\\
3\ln{\left|\frac{1-\sin{\left(t\right)}}{\cos\left(t\right)}\right|}+3t+C_{1}\\
\cos{\left(t\right)}=\sqrt[3]{x}\\
t=\arccos{\left(\sqrt[3]{x}\right)}\\
\sin{\left(t\right)}=\sqrt{1-\sqrt[3]{x^2}}\\
3\ln{\left|\frac{1-\sqrt{1-\sqrt[3]{x^2}}}{\sqrt[3]{x}}\right|}+3\arccos{\left(\sqrt[3]{x}\right)}+C\\
\)