calki na kolokwium

[estebeste]

witam, mam pare calek do rozwiazania

\(\displaystyle{ \int_{}^{} arccosx dx}\)

\(\displaystyle{ \int_{}^{} lnx dx}\)

\(\displaystyle{ \int_{}^{} \frac{ x^{2} }{x ^{2}-4x+20 }dx}\)

\(\displaystyle{ \int_{}^{} \frac{e ^{x} }{e ^{2x}+1}dx}\)

\(\displaystyle{ \int_{}^{} \frac{x ^{2} }{ \sqrt{x ^{3}+5 } }dx}\)

\(\displaystyle{ \int_{}^{} \frac{ x^{2}+2x }{x ^{2}+6x+18 }dx}\)

[Maniek]

\(\displaystyle{ \int ln(x)dx=\left[\begin{array}{c&c}u=\ln(x) & u^\prime=1/x \\ v^\prime=1 & v=x \end{array}\right]=xln(x)-\int{1dx}=xln(x)-x+C}\)

[agulka1987]

witam, mam pare calek do rozwiazania

\(\displaystyle{ \int_{}^{} arccosx dx}\)

\(\displaystyle{ =\begin{bmatrix}f'=1& g=arccosx\\f=x& g'= \frac{-1}{ \sqrt{1-x^2} } \end{bmatrix} = xarccosx + \int \frac{x}{ \sqrt{1-x^2} } dx = \begin{bmatrix}t=1-x^2\\dt=-2xdx\\- \frac{dt}{2}=xdx\end{bmatrix} = xarccosx - \frac{1}{2} \int \frac{1}{ \sqrt{t} }dt = xarccosx - \sqrt{t} +C = xarccosx - \sqrt{1-x^2} +C}\)

\(\displaystyle{ \int_{}^{} \frac{e ^{x} }{e ^{2x}+1}dx}\)

\(\displaystyle{ = \int \frac{e^x}{(e^x)^2 + 1}dx = \begin{bmatrix}t=e^x\\dt=e^x\end{bmatrix} = \int \frac{1}{t^2 + 1 } dt = arctgt+C = arctge^x+C}\)

\(\displaystyle{ \int_{}^{} \frac{x ^{2} }{ \sqrt{x ^{3}+5 } }dx}\)

\(\displaystyle{ = \begin{bmatrix} t=x^3+5\\dt=3x^2dx\\ \frac{dt}{3}=x^2dx\end{bmatrix} = \frac{1}{3} \int \frac{1}{ \sqrt{t} }dt = \frac{2}{3} \sqrt{t} +C = \frac{2}{3} \sqrt{x^3 + 5} +C}\)

[estebeste]

dzieki

[gufox]

\(\displaystyle{ \int_{}^{} \frac{ x^{2}+2x }{x ^{2}+6x+18 }dx}\)

\(\displaystyle{ ...=\int \frac{x ^{2}+6x+18-4x-18 }{x ^{2}+6x+18 }dx=\int dx - \int \frac{4x+18}{x ^{2}+6x+18 }= x-ln|x ^{2}+6x+18|-6 \int \frac{dx}{(x+3) ^{2}+3 ^{2} } = x-2ln|x ^{2}+6x+18|-2arctg \frac{x+3}{3}+C}\)