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

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

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

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

arctan (x) - x + tan (x) + C

Solution steps

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

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

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

= arctan (x) - x + tan (x)

Add a constant to the solution

= arctan (x) - x + tan (x) + C

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