Obliczyć całkę stosując podstawienie

[rwefhweo]

a) ∫\(\frac{lnx}{x(lnx-2)^2}\)

b) ∫\(\frac{ln(x+1)-lnx}{x(x+1)}\)

c) ∫\(\frac{x^5}{1+x^{12}}\)

d) ∫\(\frac{sinxcosx}{1+sin^4x}\)

e) ∫\(\frac{3^x}{1+9^x}\)

f) ∫\(\frac{dx}{e^x+e^{-x}}\)

[eresh]

a) ∫\(\frac{lnx}{x(lnx-2)^2}\)

\(\int\frac{\ln xdx}{x(\ln x-2)^2}= \begin{bmatrix} \ln x-2=t\\ \frac{dx}{x}=dt\end{bmatrix} =\int\frac{t+2dt}{t^2}=\int\frac{dt}{t}+\int\frac{2dt}{t^2}=\ln t-\frac{2}{t}+C=\ln|\ln x-2|-\frac{2}{\ln|x-2|}+C\)

[eresh]

b) ∫\(\frac{ln(x+1)-lnx}{x(x+1)}\)

\(\int\frac{ln(x+1)-lnx}{x(x+1)}dx=\int\frac{\ln\frac{x+1}{x}}{x(x+1)}dx= \begin{bmatrix}\ln\frac{x+1}{x}=t\\ \frac{-dx}{x(x+1)}=dt \end{bmatrix}=\int\frac{-tdt}{1}=-\frac{t^2}{2}+C=\frac{-(\ln\frac{x+1}{x})^2}{2}+C \)

[eresh]

c) ∫\(\frac{x^5}{1+x^{12}}\)

\(\int\frac{x^5dx}{1+x^{12}}dx= \begin{bmatrix} x^6=t\\6x^5dx=dt\end{bmatrix}=\int\frac{dt}{6(1+t^2)}=\frac{1}{6}\arctg t+C=\frac{1}{6}\arctg x^6+C\)

[eresh]

d) ∫\(\frac{sinxcosx}{1+sin^4x}\)

\(\int\frac{\sin x\cos xdx}{1+\sin^4x}= \begin{bmatrix}\sin^2x=t\\2\sin x\cos xdx=dt\end{bmatrix}=\int\frac{dt}{2(1+t^2)}=\frac{1}{2}\arctg t+C=\frac{1}{2}\arctg\sin^2x+C \)

[eresh]

e) ∫\(\frac{3^x}{1+9^x}\)

\(\int\frac{3^x}{1+9^x}dx= \begin{bmatrix}3^x=t\\3^x\ln 3dx=dt \end{bmatrix} =\int\frac{dt}{\ln 3(1+t^2)}=\frac{\arctg t}{\ln 3}+C=\frac{\arctg 3^x}{\ln 3}+C\)

[eresh]

f) ∫\(\frac{dx}{e^x+e^{-x}}\)

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