Evaluate the definite integral $\int_0^1 \frac{1}{(1 + \sqrt{x})^{4}} dx$ In chapter 4, section 5 of the 7th edition of James Stewart's Calculus book is the following problem.
51.) Evaluate $\displaystyle{\int_{0}^{1} \frac{1}{(1 + \sqrt{x})^4} \, dx}$.
In this section, only the Substitution Rule was presented.
Substitution Rule for Definite Integrals
If $g^{\prime}$ is continuous on the interval $[a, b]$ and $f$ is continuous on the image under $g$ of $[a, b]$,
\begin{equation}
\int_{a}^{b} f\bigl(g(x)\bigr) g^{\prime}(x) \, dx
= \int_{g(a)}^{g(b)} f(x) \, dx
.
\end{equation}
What solution is intended by Stewart using this rule?
I understand that with the substitution $x = t^{2}$, the definite integral can be evaluated using Integration by Parts.  This is not presented until Chapter 7 in this textbook.
This problem is not in the section on the Substitution Rule in the previous edition of the textbook.  I suspect that the problem was misplaced in the 7th edition.
 A: Let $u=1+\sqrt{x}$, so $x=(u-1)^2$ and $dx=2(u-1)du$.
Then $\displaystyle\int_0^1\frac{1}{(1+\sqrt{x})^4}dx=\int_1^2\frac{1}{u^4}\cdot2(u-1)du=2\int_1^2(u^{-3}-u^{-4})du=2\left[-\frac{1}{2}u^{-2}+\frac{1}{3}u^{-3}\right]_1^2$
$\displaystyle=2\left(-\frac{1}{8}+\frac{1}{24}-\big(-\frac{1}{2}+\frac{1}{3}\big)\right)=2\left(\frac{1}{12}\right)=\frac{1}{6}$

The Substitution Rule as stated does not apply to this example, since $g^{\prime}(x)=\frac{1}{2\sqrt{x}}$ is not continuous on $[0,1]$.
However, if $f(u)=2(u^{-3}-u^{-4}),\;\;$ then $\displaystyle f(g(x))g^{\prime}(x)=\frac{1}{(1+\sqrt{x})^4}$ for $x>0$; 
so it  is continuous on $[0,1]$ if we define its value at 0 to be 1, 
and therefore the conclusion of the Substitution Rule still applies. 
A: Replace $x$ with $u^2$ to get $dx=2u\,du$ then:
$$ I = 2\int_{0}^{1}\frac{u}{(1+u)^4}\,du = 2\int_{0}^{1}\frac{du}{(1+u)^3}-2\int_{0}^{1}\frac{du}{(1+u)^4}, $$
from which:
$$ I = 2\cdot\left[-\frac{1}{2}\cdot\frac{1}{(1+u)^2}+\frac{1}{3}\cdot\frac{1}{(1+u)^3}\right]_{0}^{1} = 2\cdot\frac{1}{12}=\color{red}{\frac{1}{6}}.$$
A: I think this may be what you mean, without partial fractions, integration by parts or other more advanced methods:
Substitute $\;u=1+\sqrt x\implies du=\frac{dx}{2\sqrt x}\implies dx=2(u-1)\,du\;$ , and
$$\int_0^1\frac{dx}{(1+\sqrt x)^4}=\int_1^2\frac{2(u-1)\,du}{u^4}=2\int_1^2\frac{du}{u^3}-2\int_1^2\frac{du}{u^4}=$$
$$=\left.-\frac1{u^2}\right|_1^2+\left.\frac23\frac1{u^3}\right|_1^2=-\frac14+1+\frac2{24}-\frac23=\frac16$$
A: Try the change of variables $x(t) := t^2$ that maps $[0,1]$ to itself. It is positive, bijective and continuously differentiable, and yields
$$
\int_{0}^1 \frac{\mathrm{d}x}{(1+\sqrt x)^4} = \int_0^1 \frac{2t}{(1+t)^4}\mathrm d t
$$
which can be readily integrated by parts, or computed straightaway by noticing that $t = 1 + t -1$.
