Integral of a hyperbolic function

Question

$$\int \tanh(x) - \tanh^3(x)\,dx$$

I get the answer as $\tanh x + c$?

I took out a factor of $\tanh x$, used the identity $1-\tanh^2 x=\text{sech}^2x$, used the substitution of $u=\tanh x$,

and reduced the question to $\int 1 du$.

Is this the way you would normally approach this question?

Cheers.

Gurjinder.B

Answer 1

The derivative of $f(x)=\tanh x$ is $$ f'(x)=\frac{\cosh^2x-\sinh^2x}{\cosh^2x}=1-\tanh^2x $$ so the integral is $$ \int f(x)f'(x)\,dx=\frac{1}{2}(f(x))^2+c=\frac{1}{2}\tanh^2x+c $$