What is the integral of+x^2sec^2(x^3+1) ?

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

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integral of+x^2sec^2(x^3+1)

\int + x^{2} sec^{2} (x^{3} + 1) d x

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

Solution steps

\int x^{2} sec^{2} (x^{3} + 1) d x

Apply u-substitution

: \int \frac{sec ^{2} ( u )}{3} d u

= \int \frac{sec ^{2} ( u )}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{3} tan (u)

Substitute back

u = x^{3} + 1

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{3} (1 + y) + z^{2})

\frac{d}{d x} (\frac{x ^{2}}{( 5 + x )})

arctan (x) - \frac{π}{4}

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

x \to 35 lim (\frac{x ^{2}}{5} - \frac{7}{x})

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