When would u-substitution be used in the numerator?

Question

For most integrals I have came across, u is almost always substituted in the denominator. However, I came across the following integral: $$\int\frac{\sqrt{x}}{1+x}dx$$

I intuitively thought that $1+x$ would be substituted, but according to the solutions, u had to be ${\sqrt{x}}$.

My question is, how do I distinguish between when u has to be substituted in the denominator versus the numerator?

Answer 1

$u=1+x$ is fine. Your integral changes to

$$\int \dfrac{\sqrt{u-1}}{u} du$$

Here you find yourself in need of another substitution though, for example $u=\sec^2t$. Note that this makes $t=\sec^{-1}(\sqrt{x+1})$.

This gives

$$\int \dfrac{\tan t}{\sec^2 t}\cdot 2\sec^2 t\tan t \ dt=2\int \tan^2 t \ dt=2(\tan t-t) + c$$

Back substituting gives the answer

$$2\tan(\sec^{-1}(\sqrt{x+1}))-2\sec^{-1}(\sqrt{x+1})+c$$

This can be simplified further using trigonometric properties if desired.

Answer 2

$$I= \int \frac{\sqrt{x} dx}{1+x}$$ Your substitution should yield integral of $u$ which is doable. Hereif you take $x=u^2 \implies dx=2udu$,then $$I=\int\frac{2u^2du}{1+u^2}=2 \int \left( 1-\frac{1}{1+u^2}\right)du=2 [u-\tan^{-1} u]=2[\sqrt{x}-\tan^{-1} \sqrt{x}]+C.$$

Answer 3

Hyperbolic substitution

Letting $x=\sinh ^2 \theta,$ then $d x=2 \sinh \theta \cosh \theta d \theta$ transforms the integral to $$ \begin{aligned} I &=\int \frac{\sinh \theta}{1+\sinh ^2 \theta} \cdot 2 \sinh \theta \cos \theta d \theta \\ &=2 \int \frac{\sinh ^2 \theta}{\operatorname{coch}_\theta} d \theta \\ &=2 \int \frac{\cosh ^2 \theta-1}{\cosh \theta} d \theta \\ &=2 \int(\cosh \theta-\operatorname{sech} \theta) d \theta \\ &=2\left[\sinh \theta-\tan ^{-1}(\sinh \theta)\right]+C\\&= 2\left(\sqrt{x}-\tan ^{-1} \sqrt{x}\right)+C \end{aligned} $$