Całki

[PriCe]

Obliczyć Całki:

a)\(\int(4x^6- \frac{4}{x^5} ) dx\)
b)\(\int\frac{(x^2+3)^2}{x} dx\)
c)\(\int \frac{dx}{5x-3}\)
d)\(\int \sqrt{6x-7} dx\)
e)\(\int x^2sinxdx\)
f)\(\int e^xsinxdx\)
g)\(\int x^2lnxdx\)

Oblicz Całkę Oznaczoną

\(\int_{0}^{2} (6x-5)^{-3}dx\)

[irena]

a)
\(\int(4x^6-\frac{4}{x^5})dx=4\int\ x^6dx-4\int\ x^{-5}dx=4\cdot\frac{1}{7}x^7-4\cdot\frac{1}{-4}x^{-4}+C=\frac{4}{7}x^7+x^{-4}+C\)

[irena]

b)
\(\int\frac{(x^2+3)^2}{x}dx=\int\frac{x^4+6x^2+9}{x}dx=\int\ x^3dx+6\int\ xdx+9\int\frac{dx}{x}=\\=\frac{1}{4}x^4+6\cdot\frac{1}{2}x^2+9ln|x|+C=\frac{1}{4}x^4+3x^2+9ln|x|+C\)

[irena]

c)
\(\int\frac{dx}{5x-3}=\\ \left(5x-3=t\\5dx=dt\\dx=\frac{1}{5}dt \right) \\=\int\frac{\frac{1}{5}dt}{t}=\frac{1}{5}\int\frac{dt}{t}=\frac{1}{5}ln|t|+C=\frac{1}{5}ln|5x-3|+C\)

[irena]

d)
\(\int\sqrt{6x-7}dx=\\ \left(6x-7=t\\6dx=dt\\dx=\frac{1}{6}dt \right) \\=\int\sqrt{t}\cdot\frac{1}{6}dt=\frac{1}{6}\int\ t^{\frac{1}{2}}dt=\frac{1}{6}\cdot\frac{1}{\frac{3}{2}}t\sqrt{t}+C=\frac{1}{9}(6x-7)\sqrt{6x-7}+C\)

[irena]

e)
\(\int\ x^2sinxdx=\\ \left(u=x^2\\u'=2x\\v'=sinx\\v=-cosx \right) \\=-x^2cosx-\int2x(-cosx)dx=-x^2cosx+2\int\ xcosxdx=\\ \left(f=x\\f'=1\\g'=cosx\\g=sinx \right) \\=-x^2cosx+2(xsinx-\int\ sinxdx)=-x^2cosx+2xsinx-2(-cosx)+C=\\=-x^2cosx+2xsinx+2cosx+C\)

[irena]

f)
\(\int\ e^xsinxdx=\\ \left(u=e^x\\u'=e^x\\v'=sinx\\v=-cosx \right) \\=-e^xcosx-\int(-e^xcosx)dx=-e^x+\int\ e^xcosxdx=\\ \left(f=e^x\\f'=e^x\\g'=cosx\\g=sinx \right) \\=-e^xcosx+(e^xsinx-\int\ e^xsinx)=e^x(sinx-cosx)-\int\ e^xsinxdx\\2\int\ e^xsinxdx=e^x(sinx-cosx)\\\int\ e^xsinx=\frac{e^x(sinx-cosx)}{2}+C\)

[irena]

g)
\(\int\ x^2lnxdx=\\ \left(u=lnx\\u'=\frac{1}{x}\\v'=x^2\\v=\frac{1}{3}x^3 \right) \\=\frac{1}{3}x^3lnx-\int\frac{1}{x}\cdot\frac{1}{3}x^3dx=\frac{1}{3}x^3lnx-\frac{1}{3}\int\ x^2dx=\frac{1}{3}x^3lnx-\frac{1}{3}\cdot\frac{1}{3}x^3+C=\frac{x^3(3lnx-1)}{9}+C\)

[PriCe]

irena zobacz pw

[irena]

\(\int(6x-5)^{-3}dx=\\ \left(6x-5=t\\6dx=dt\\dx=\frac{1}{6}dt \right) \\=\int\ t^{-3}\cdot\frac{1}{6}dt=\frac{1}{6}\cdot\frac{1}{-2}t^{-2}+C=-\frac{1}{12(6x-5)^2}+C\)

\(\int_0^2(6x-5)^{-3}=[-\frac{1}{12(6x-5)^2}]_0^2=\\=(-\frac{1}{12(12-5)^2})-(-\frac{1}{12(0-5)^2})=\\=-\frac{1}{12\cdot49}+\frac{1}{12\cdot25}=\\=\frac{49-25}{12\cdot25\cdot49}=\frac{24}{12\cdot49\cdot25}=\frac{2}{1225}\)