Calculate area under curve $(y+1)^2 x=1$ in first quadrant

Question

$$(y+1)^2 x=1$$

We have to calculate area under this curve, bounded by ordinate and abscissa in first quadrant. Task also says that we have to express area as improper integral and then calculate it.

This is what i got so far.

$y=\sqrt{\frac1{x}}-1$

I think area is given by $$\int_{0\leftarrow}^{+\infty} \left ( \sqrt{\frac1{x}}-1\right )dx=\int_{0\leftarrow}^{+\infty}x^{-\frac1{2}}dx\quad -\int_{0\leftarrow}^{+\infty}dx$$

I get some weird stuff when solving this integrals, so do you think I am on right path?

Answer 1

The graph meets the $x$-axis at $x=1$, after which the curve remains below the $x$-axis. So, integrating your expression from $0$ to $1$ is sufficient.

$$\text{Area}=\int_0^1\left(\frac{1}{\sqrt{x}}-1\right)dx$$

Answer 2

Alternatively, you could view $x$ as a function of $y$ and hence write $x = \frac{1}{(y+1)^2}$. Now consider the integral $\int_0^{\infty} \frac{1}{(y+1)^2} \, dy$.

Answer 3

Your upper limit of integration seems to be incorrect as the function cuts x axis at 1