What is the integral of sec(x)(tan(x)+sec(x)) ?

The integral of sec(x)(tan(x)+sec(x)) is sec(x)+tan(x)+C

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integral of sec(x)(tan(x)+sec(x))

\int sec (x) (tan (x) + sec (x)) d x

sec (x) + tan (x) + C

Solution steps

\int sec (x) (tan (x) + sec (x)) d x

Rewrite using trig identities

= \int \frac{tan ( x ) cos ( x ) + 1}{cos ^{2} ( x )} d x

Expand

\frac{tan ( x ) cos ( x ) + 1}{cos ^{2} ( x )} : \frac{tan ( x )}{cos ( x )} + \frac{1}{cos ^{2} ( x )}

= \int \frac{tan ( x )}{cos ( x )} + \frac{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 \frac{tan ( x )}{cos ( x )} d x + \int \frac{1}{cos ^{2} ( x )} d x

\int \frac{tan ( x )}{cos ( x )} d x = sec (x)

\int \frac{1}{cos ^{2} ( x )} d x = tan (x)

= sec (x) + tan (x)

Add a constant to the solution

= sec (x) + tan (x) + C

\int \frac{1}{1 + e ^{2 x}} d x

y^{^{'}} = a y + b e^{- b}

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

y = \frac{4}{( 7 x ) ^{7}}

\int \frac{1}{x ^{2} 64 x ^{2} + 25} d x

What is the integral of sec(x)(tan(x)+sec(x)) ?
The integral of sec(x)(tan(x)+sec(x)) is sec(x)+tan(x)+C