Does this one require integration by parts?

Question

$$\int_{0}^{1} x \sqrt{1+8x^2}\,dx$$

I tried

$u=x$

$\text{d}u=\text{d}x$

$\text{d}v = (1+8x^2)^\frac{1}{2}$

but I'm not sure how to get $v$ by integrating $dv$, since $u$-sub doesn't work.

Barking up the wrong tree?

Answer 1

Hint:

Substitute:

$$ 1+8x^2=u \quad \Rightarrow \quad 16xdx=du \quad \Rightarrow \quad x dx=\frac{1}{16}du $$

Answer 2

(Joke) We will use integration by parts!

Let $u=\sqrt{1+8x^2}$ and $dv=x\,dx$. Then $du=\frac{8x}{\sqrt{1+8x^2}}\,dx$ and we can let $v=\frac{x^2}{2}+\frac{1}{16}=\frac{1+8x^2}{16}$. Thus our integral $I$ is given by $$I=\left.\frac{1}{16}(1+8x^2)^{3/2}\large\right|_0^1-\int_0^1 \frac{1}{2}x\sqrt{1+8x^2}\,dx=\left.\frac{1}{16}(1+8x^2)^{3/2}\large\right|_0^1-\frac{I}{2}.$$ It follows that $$\frac{3}{2}I=\frac{1}{16}(27-1),$$ so $I=\frac{13}{12}$.

Remark: After learning about integration by parts, some students develop a great fondness for it, and may miss substitutions that would have been routine a few days earlier.

Answer 3

With the not so-trivial-substitution $x=\frac{\sinh(z)}{2\sqrt{2}}$ the integral becomes:

$$ \frac{1}{8}\int_{0}^{\text{arcsinh}(2\sqrt{2})}\sinh(u)\cosh(u)^2\,du=\left.\frac{\cosh^3(u)}{24}\right|_{0}^{\text{arccosh}(3)}=\frac{27-1}{24}=\color{red}{\frac{13}{12}}.$$