$\int h(x)h'(x) dx$ without u-substitution/parts

Question

I have to calculate the following antiderivative without using u-sub, even though it's very easy that way. Any suggestions?

$$\int h(x)h'(x) dx$$

Answer 1

There is no need to use “rules” for computing an antiderivative. Since the derivative of $$ f(x)=\frac{1}{2}(h(x))^2 $$ is $$ f'(x)=h(x)h'(x) $$ by the chain rule, you're done: an antiderivative of $h(x)h'(x)$ (over an interval) has the form $$ \frac{1}{2}(h(x))^2+c $$ for some constant $c$.

Answer 2

Use integrating by parts to find an expression of $\int h(x) h'(x) \text{d}x$.

Answer 3

By parts, let $u=h(x)$, $dv=h'(x)dx$ Then $$\int h(x)h'(x)dx=h^2(x)+C-\int h(x)h'(x)dx$$ $$I=h^2(x)+C-I$$ So $$I=\frac{1}{2}h^2(x)+C$$

Answer 4

$\int h(x)h'(x)dx$

Set $u=h(x)$ and $dv=h'(x)dx$ then $$ du=h'(x)dx $$ and $$ v=h(x) $$ Then by the formula $uv-\int v du$ this is $$ h(x)^2-\int h(x)h'(x)dx $$ Thus $$ \int h(x)h'(x)dx=h(x)^2-\int h(x)h'(x)dx $$ So $$ 2\int h(x)h'(x)dx=h(x)^2 $$ so $$ \int h(x)h'(x)dx=\frac{h(x)^2}{2}+C $$

Answer 5

You can do integration by parts this gives,

$$I=\int h(x)h'(x) dx$$

$$=h(x)h(x)-\int h(x)h'(x) dx$$

$$=h^2(x)-I+C_1$$

So that,

$$I=\frac{1}{2}h^2(x)+C$$