What is the integral from 1 to 2 of (3t^2+2e^{3t}+1/t+2cos(3t)) ?

The integral from 1 to 2 of (3t^2+2e^{3t}+1/t+2cos(3t)) is ln(2)+(2(e^6-e^3)+2(sin(6)-sin(3)))/3+7

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integral from 1 to 2 of (3t^2+2e^{3t}+1/t+2cos(3t))

\int_{1}^{2} (3 t^{2} + 2 e^{3 t} + \frac{1}{t} + 2 cos (3 t)) d t

ln (2) + \frac{2 ( e ^{6} - e ^{3} ) + 2 ( sin ( 6 ) - sin ( 3 ) )}{3} + 7

Decimal

262.97496 \dots

Solution steps

\int_{1}^{2} 3 t^{2} + 2 e^{3 t} + \frac{1}{t} + 2 cos (3 t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 3 t^{2} d t + \int_{1}^{2} 2 e^{3 t} d t + \int_{1}^{2} \frac{1}{t} d t + \int_{1}^{2} 2 cos (3 t) d t

\int_{1}^{2} 3 t^{2} d t = 7

\int_{1}^{2} 2 e^{3 t} d t = \frac{2 ( e ^{6} - e ^{3} )}{3}

\int_{1}^{2} \frac{1}{t} d t = ln (2)

\int_{1}^{2} 2 cos (3 t) d t = \frac{2}{3} (sin (6) - sin (3))

= 7 + \frac{2 ( e ^{6} - e ^{3} )}{3} + ln (2) + \frac{2}{3} (sin (6) - sin (3))

Simplify

7 + \frac{2 ( e ^{6} - e ^{3} )}{3} + ln (2) + \frac{2}{3} (sin (6) - sin (3)) : ln (2) + \frac{2 ( e ^{6} - e ^{3} ) + 2 ( sin ( 6 ) - sin ( 3 ) )}{3} + 7

= ln (2) + \frac{2 ( e ^{6} - e ^{3} ) + 2 ( sin ( 6 ) - sin ( 3 ) )}{3} + 7

What is the integral from 1 to 2 of (3t^2+2e^{3t}+1/t+2cos(3t)) ?
The integral from 1 to 2 of (3t^2+2e^{3t}+1/t+2cos(3t)) is ln(2)+(2(e^6-e^3)+2(sin(6)-sin(3)))/3+7