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

The integral of 1/3 x^2tanh(x^3+1) is 1/9 ln|cosh(x^3+1)|+C

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integral of 1/3 x^2tanh(x^3+1)

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

\frac{1}{9} ln cosh (x^{3} + 1) + C

Solution steps

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

Take the constant out:

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

= \frac{1}{3} \cdot \int x^{2} tanh (x^{3} + 1) d x

Apply u-substitution

: \int \frac{tanh ( u )}{3} d u

= \frac{1}{3} \cdot \int \frac{tanh ( 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 \frac{1}{3} \cdot \int tanh (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int \frac{sinh ( u )}{cosh ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{3} \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} ln cosh (x^{3} + 1)

Simplify

\frac{1}{3} \cdot \frac{1}{3} ln cosh (x^{3} + 1) : \frac{1}{9} ln cosh (x^{3} + 1)

= \frac{1}{9} ln cosh (x^{3} + 1)

Add a constant to the solution

= \frac{1}{9} ln cosh (x^{3} + 1) + C

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What is the integral of 1/3 x^2tanh(x^3+1) ?
The integral of 1/3 x^2tanh(x^3+1) is 1/9 ln|cosh(x^3+1)|+C