What is the integral from 0 to pi/4 of 1+tan(x) ?

The integral from 0 to pi/4 of 1+tan(x) is pi/4+1/2 ln(2)

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integral from 0 to pi/4 of 1+tan(x)

\int_{0}^{\frac{π}{4}} 1 + tan (x) d x

\frac{π}{4} + \frac{1}{2} ln (2)

Decimal

1.13197 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 1 + tan (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_{0}^{\frac{π}{4}} 1 d x + \int_{0}^{\frac{π}{4}} tan (x) d x

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

\int_{0}^{\frac{π}{4}} tan (x) d x = \frac{1}{2} ln (2)

= \frac{π}{4} + \frac{1}{2} ln (2)

\frac{d}{d x} (\frac{6 x}{x ^{2} - 9})

y^{^{'}} = (3 x - 2) (30 x + 29)

\frac{d y}{d x}, 9 x^{2} + y^{2} = 1

x \to - \infty lim (x^{2})

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 12 x^{2}

What is the integral from 0 to pi/4 of 1+tan(x) ?
The integral from 0 to pi/4 of 1+tan(x) is pi/4+1/2 ln(2)