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

The integral of xsec(3x^2+1) is 1/6 ln|tan(3x^2+1)+sec(3x^2+1)|+C

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integral of xsec(3x^2+1)

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

\frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{sec ( 3 u + 1 )}{2} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int sec (v) \frac{1}{3} d v

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

Take the constant out:

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

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

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} \cdot \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

Simplify

\frac{1}{2} \cdot \frac{1}{3} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) : \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

= \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1)

Add a constant to the solution

= \frac{1}{6} ln tan (3 x^{2} + 1) + sec (3 x^{2} + 1) + C

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What is the integral of xsec(3x^2+1) ?
The integral of xsec(3x^2+1) is 1/6 ln|tan(3x^2+1)+sec(3x^2+1)|+C