Question #110775
Find the indefinite integral of the following functions
Y= 3t^2+2e^3t+1/t+ cos 3t
Y=x/2+3x^2
1
Expert's answer
2020-04-27T18:51:18-0400

(3t2+2e3t+1/t+cos3t)dt=3t2dt+2e3tdt+dt/t+cos(3t)dt=∫(3t^2+2e^{3t} +1/t+cos3t)dt=3∫t^2 dt+2∫e^{3t}dt +∫dt/t+∫cos(3t)dt= =3(1/3)t3+2(1/3)3t+lnt+(1/3)sin(3t)+C=t3+(2/3)3t+lnt+(1/3)sin(3t)+C=3⋅(1/3) t^3+2⋅(1/3) ⅇ^{3t}+ln⁡|t|+(1/3) sin⁡(3t)+C=t^3+(2/3) ⅇ^{3t}+ln⁡|t|+(1/3) sin(⁡3t)+C\\


2)(x/2+3x2)dx=(1/2)xdx+3x2dx=\\2) ∫( x/2+3x^2)dx=(1/2) ∫xdx+3∫x^2 dx=

=(1/2)(1/2)x2+3(1/3)x3+C=(1/4)x2+x3+C=(1/2)⋅(1/2) x^2+3⋅(1/3) x^3+C=(1/4) x^2+x^3+C


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