What is the integral from 1 to 4 of 3/(2x+1)+cos(2x) ?

The integral from 1 to 4 of 3/(2x+1)+cos(2x) is (3ln(3)+sin(8)-sin(2))/2

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integral from 1 to 4 of 3/(2x+1)+cos(2x)

\int_{1}^{4} \frac{3}{2 x + 1} + cos (2 x) d x

\frac{3 ln ( 3 ) + sin ( 8 ) - sin ( 2 )}{2}

Decimal Notation

1.68794 \dots

Solution steps

\int_{1}^{4} \frac{3}{2 x + 1} + cos (2 x) d x

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}^{4} \frac{3}{2 x + 1} d x + \int_{1}^{4} cos (2 x) d x

\int_{1}^{4} \frac{3}{2 x + 1} d x = \frac{3}{2} ln (3)

\int_{1}^{4} cos (2 x) d x = \frac{1}{2} (sin (8) - sin (2))

= \frac{3}{2} ln (3) + \frac{1}{2} (sin (8) - sin (2))

Simplify

\frac{3}{2} ln (3) + \frac{1}{2} (sin (8) - sin (2)) : \frac{3 ln ( 3 ) + sin ( 8 ) - sin ( 2 )}{2}

= \frac{3 ln ( 3 ) + sin ( 8 ) - sin ( 2 )}{2}

\int_{1}^{2} (e^{x} + \frac{1}{x}) d x

\int_{- 1}^{x} \frac{1}{1 + t ^{2}} d t

\int_{0}^{x} t e^{t^{2}} d t

\int_{0}^{1} cos (x + y) d x

\int_{1}^{3} \frac{x ^{4} + 1}{x ^{2}} d x

What is the integral from 1 to 4 of 3/(2x+1)+cos(2x) ?
The integral from 1 to 4 of 3/(2x+1)+cos(2x) is (3ln(3)+sin(8)-sin(2))/2