Why do you need to perform integration by substitution when integrating $\frac{1}{1+\sqrt{x}}$?

Question

For the function $\frac{1}{1+\sqrt{x}}$, typically the correct approach is to perform integration by substitution like so:

$$ \int \frac{1}{1+\sqrt{x}} \, \mathrm{d}x $$

Let $u = 1 + \sqrt{x}$, i.e., $x=(u-1)^2$. Then $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{2\sqrt{x}}$, or equivalently, $\mathrm{d}x = 2\sqrt{x} \, \mathrm{d}u$. Substituting in, we get:

$$ 2\int \frac{u-1}{u} \, \mathrm{d}u =2 \left( 1+\sqrt{x} - \ln(1+\sqrt{x}) \right). $$

I then proceeded to try the same integration using a sort of reverse chain rule:

$$ \int \frac{1}{1+\sqrt{x}} \, \mathrm{d}x = \ln(1+\sqrt{x}) \cdot \frac{1}{1/(2\sqrt{x})} = \ln(1+\sqrt{x}) \cdot 2\sqrt{x}. $$

However this is clearly incorrect. At school we are just taught that when doing integration of the style $\frac{1}{x+k}$, we need to perform integration by substitution but why is this?

For example, if I were to integrate $\frac{1}{2x+1}$, using reverse chain rule method, it achieves the correct result of $\frac{1}{2}\ln(1+\sqrt{x})$?My first thought is it has something to do with differentiation of $\mathrm{d}u/\mathrm{d}x$ that gives a function of $x$.

Answer 1

Integration by substitution is actually the chain rule in reverse.

If you can find some functions $f$ and $g$ such that the integrand can be expressed as $g'(x)f(g(x))$ then you can integrate it as $F(g(x))$.

Say you suppose that $g(x) = \sqrt{x}$, then $g'(x) = 0.5/\sqrt{x}$. Then we write

$$\frac{1}{1+\sqrt{x}} = \frac{0.5/\sqrt{x}}{0.5/\sqrt{x}} \frac{1}{1+\sqrt{x}} = (0.5/\sqrt{x}) \frac{2\sqrt{x}}{(1+\sqrt{x})} $$

The integral of $f(x) = \frac{x}{1+x} = 1 - \frac{1}{1+x}$ we recognize as $x - \log(1+x)$

By performing the substitution it will be more easy to directly find what the form of the formula $f$ is.

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Your application of the chain rule directly might also work but it depends on a remainder term and whether it can be eliminated.

Your idea is to integrate $\int f(g(x)) dx$ by taking $F(g(x))/g'(x)$ but that is not entirely correct since the reverse process would need the product rule

$$\frac{\partial}{\partial x} \left( F(g(x)) \cdot \frac{1}{g'(x) } \right) = f(g(x)) - F(g(x)) \cdot \frac{g''(x)}{g'(x)^2}$$

So it is almost correct. You have this additional term $F(g(x)) \cdot \frac{g''(x)}{g'(x)^2}$. If that term is easy and you can integrate this then you could correct it and use

$$\frac{\partial}{\partial x} \left( F(g(x)) \cdot \frac{1}{g'(x) } + \int F(g(x)) \cdot \frac{g''(x)}{g'(x)^2} dx \right) = f(g(x)) $$

Answer 2

Method 1

Let $y=1+\sqrt{x}$, then $x=(y-1)^{2}$, then $d x =2(y-1) d y $ and

$$ \begin{aligned} I &=\int \frac{1}{y} \cdot 2(y-1) d y \\ &=2 \int\left(1-\frac{1}{y}\right) d y \\ &=2 y-2 \ln |y|+C_{1} \\ &=2(1+\sqrt{x})-2 \ln |1+\sqrt{x}|+C_{1} \\ &=2 \sqrt{x}-2 \ln |1+\sqrt{x}|+C_{2} \end{aligned} $$ Method 2

Let $y=\sqrt{x}$, then $y^{2}=x \Rightarrow d x=2 y d y$, then $$ \begin{aligned} I &=\int \frac{1}{1+y} \cdot 2 y d y \\ &=2 \int\left(1-\frac{1}{1+y}\right) d y \\ &=2 y-2 \ln |1+y|+C . \\ &=2 \sqrt{x}-2 \ln |1+\sqrt{x}|+C \end{aligned} $$ Wish it helps.