What is the integral of (sec^2(t)i+1/(1+t^2)j) ?

The integral of (sec^2(t)i+1/(1+t^2)j) is itan(t)+jarctan(t)+C

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integral of (sec^2(t)i+1/(1+t^2)j)

\int (sec^{2} (t) i + \frac{1}{1 + t ^{2}} j) d t

i tan (t) + j arctan (t) + C

Solution steps

\int sec^{2} (t) i + \frac{1}{1 + t ^{2}} j d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sec^{2} (t) i d t + \int \frac{1}{1 + t ^{2}} j d t

\int sec^{2} (t) i d t = i tan (t)

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

= i tan (t) + j arctan (t)

Add a constant to the solution

= i tan (t) + j arctan (t) + C

What is the integral of (sec^2(t)i+1/(1+t^2)j) ?
The integral of (sec^2(t)i+1/(1+t^2)j) is itan(t)+jarctan(t)+C