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Integration of ln 5x dx?
5/5x + c where c is the constant of intergration just differentiate the 5x to get 5 and times that by 1/5x then add c The answer above is wrong. This is simply because you cannot intergrate lnx to get 1/x. To intergrate I would recommend using intergration by parts. u=ln5x du/dx=1/x dv/dx=1 v=x uv-(intergal of)v.du/dx =xln5x-intergral of x/x intergral of x/x = x =xln5x-x+c = x(ln5x-1)+c
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
What is the derivative of ln 1 divided by 1-x?
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What is the derivative of sin x to the e to the xth power?
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
Integration of tangent x?
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
