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

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

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integral of sec^2(x)+1

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

tan (x) + x + C

Solution steps

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

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

\int 1 d x = x

= tan (x) + x

Add a constant to the solution

= tan (x) + x + C

\int sin^{3} (4 x) cos^{3} (4 x) d x

y^{^{'}} = (2 x + y)^{2}

y^{^{''}} + 7 y + 10 = 0, y (0) = 3, y^{^{'}} (0) = 18

\frac{d}{d s} (\frac{6 s ^{3} + 28 s ^{2} + 38 s + 15}{( s + 1 ) ^{2}})

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x - 3)