#int (3x^3+5)/(5x-3) dx=#?

1 Answer
Mar 29, 2018

#I=(x^3)/5+(9x^2)/50+(27x)/125+2118/1875 ln|5x-3|+c#

Explanation:

Here,

#I=int(3x^3+5)/(5x-3)dx#

#=3/5int(x^3+5/3)/(x-3/5)dx#

#=3/5int ((x^3-27/125+706/375)/(x-3/5))dx#

#=3/5int((x^3-(3/5)^3)/(x-3/5))dx+3/5int((706)/375)/(x-3/5)dx#

#=3/5int(cancel((x-3/5))(x^2+(3x)/5+9/25))/(cancel((x- 3/5)))dx+2118/1875int(5)/(5x-3)dx#

#=3/5int(x^2+(3x)/5+9/25)dx+2118/1875int(d/(dx)(5x-3))/(5x- 3)dx#

#=3/5[x^3/3+(3x^2)/10+(9x)/25]+2118/1875ln|5x-3|+c#

#I=(x^3)/5+(9x^2)/50+(27x)/125+2118/1875 ln|5x-3|+c#