What is the integral of (sec^2(x))(2tan(x)+1) ?

The integral of (sec^2(x))(2tan(x)+1) is tan^2(x)+tan(x)+C

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integral of (sec^2(x))(2tan(x)+1)

\int (sec^{2} (x)) (2 tan (x) + 1) d x

tan^{2} (x) + tan (x) + C

Solution steps

\int sec^{2} (x) (2 tan (x) + 1) d x

Rewrite using trig identities

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

Expand

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

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

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

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

= tan^{2} (x) + tan (x)

Add a constant to the solution

= tan^{2} (x) + tan (x) + C

\frac{\partial}{\partial x} (\frac{1}{4} x^{2} \cdot y)

\frac{d y}{d x} - \frac{4 y}{x} = \frac{y ^{4}}{x ^{2}}

\frac{\partial}{\partial x} (x e^{(\frac{x}{y})})

f (x) = (\frac{x}{2})^{4}

f (x) = (x - 1)^{\frac{2}{3}}

What is the integral of (sec^2(x))(2tan(x)+1) ?
The integral of (sec^2(x))(2tan(x)+1) is tan^2(x)+tan(x)+C