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

The integral of (2+tan^2(x))^2 is x+1/3 tan^3(x)+3tan(x)+C

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

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

x + \frac{1}{3} tan^{3} (x) + 3 tan (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( 2 cos ^{2} ( x ) + sin ^{2} ( x ) ) ^{2}}{cos ^{4} ( x )} d x

Expand

\frac{( 2 cos ^{2} ( x ) + sin ^{2} ( x ) ) ^{2}}{cos ^{4} ( x )} : 4 + \frac{4 sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{4} ( x )}{cos ^{4} ( x )}

= \int 4 + \frac{4 sin ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{4} ( x )}{cos ^{4} ( 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 4 d x + \int \frac{4 sin ^{2} ( x )}{cos ^{2} ( x )} d x + \int \frac{sin ^{4} ( x )}{cos ^{4} ( x )} d x

\int 4 d x = 4 x

\int \frac{4 sin ^{2} ( x )}{cos ^{2} ( x )} d x = 4 (tan (x) - x)

\int \frac{sin ^{4} ( x )}{cos ^{4} ( x )} d x = \frac{1}{3} tan^{3} (x) + x - tan (x)

= 4 x + 4 (tan (x) - x) + \frac{1}{3} tan^{3} (x) + x - tan (x)

Simplify

4 x + 4 (tan (x) - x) + \frac{1}{3} tan^{3} (x) + x - tan (x) : x + \frac{1}{3} tan^{3} (x) + 3 tan (x)

= x + \frac{1}{3} tan^{3} (x) + 3 tan (x)

Add a constant to the solution

= x + \frac{1}{3} tan^{3} (x) + 3 tan (x) + C

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What is the integral of (2+tan^2(x))^2 ?
The integral of (2+tan^2(x))^2 is x+1/3 tan^3(x)+3tan(x)+C