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

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

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    Hmmm...But why? – 2017-02-01
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    what can you use – 2017-02-01
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    Just open your eyes. The primitive $\frac{1}{2} h^2$ will jump in. – 2017-02-01
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    @Paul, since this question is assigned from a section of my textbook that hasnt covered u-sub yet – 2017-02-01
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    It's highly unusual to have integration by parts before u substitution. Does that work? – 2017-02-01
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    Can't do parts either, i forgot to add it in the question – 2017-02-01
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    I guess I have to use some sort of thereom to solve this – 2017-02-01
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    Do you know the fundamental theorem of calculus? – 2017-02-01
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    Yes I konw that – 2017-02-01
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    But that doesnt relaly apply here since Im not calculating a definite integral – 2017-02-01

5 Answers 5

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

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 $$ \int h(x)h'(x)dx=\frac{h(x)^2}{2}+C $$

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    I can't do parts either Dx – 2017-02-01
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    what can you use – 2017-02-01
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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,

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

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    Rats! Beaten to the punch! +1 – 2017-02-01
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    I can't do parts either Dx, shoudlve mentioned it in the question – 2017-02-01
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    Then what can you do @KSplitX ? – 2017-02-01
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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 $c$.

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    Perhaps your right, maybe its just a direct concept type question. Thanks – 2017-02-01
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Use integrating by parts to find an expression of $\int h(x) h'(x) \text{d}x$.

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