integration

[pinkcrushin22]

Integrate sqrt(x)/ (x-4) (u=sqrt(x))

u= sqrt(x), du= 1/2x/^-1/2, 2sqrt(x)du= dx, 2udu=dx

Integral u^2/u^2-2^2du

How do I solve from here?

 

[MarkFL]

We are given to integrate:

\(\displaystyle \displaystyle \int\frac{\sqrt{x}}{x-4}\,dx\)

Using the substitution:

\(\displaystyle \displaystyle u=\sqrt{x}\,\therefore\,dx=2u\,du\) and we have:

\(\displaystyle \displaystyle 2\int\frac{u^2}{u^2-2^2}\,dx=2\int\frac{(u^2-2^2)+2^2}{u^2-2^2}\,dx=2\int 1+4\frac{1}{u^2-2^2}\,dx\)

From here, I would consider partial fraction decomposition using the Heaviside cover-up method for simplicity.

 

[soroban]

Hello, pinkcrushin22!

\(\displaystyle \displaystyle\int \frac{\sqrt{x}}{x-4}\,dx\)

Let \(\displaystyle u \,=\,\sqrt{x}\quad\Rightarrow\quad x \,=\,u^2 \quad\Rightarrow\quad dx \,=\,2u\,du\)

Substitute: .\(\displaystyle \displaystyle \int \frac{u}{u^2-4}(2u\,du) \;=\;2\int\frac{u^2}{u^2-4}\,du \;=\; 2\int\left(1 + \frac{4}{u^2-4}\right)\,du \)

. . . . \(\displaystyle \displaystyle =\; 2\int du + 8 \int\frac{du}{u^2-4} \;=\;2u + \tfrac{8}{4}\ln\left|\frac{u-2}{u+2}\right| + C\)


Back-substitute: .\(\displaystyle 2\sqrt{x} + 2\ln\left|\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\right| + C\)

 

[pinkcrushin22]

helpful

We are given to integrate:

\(\displaystyle \displaystyle \int\frac{\sqrt{x}}{x-4}\,dx\)

Using the substitution:

\(\displaystyle \displaystyle u=\sqrt{x}\,\therefore\,dx=2u\,du\) and we have:

\(\displaystyle \displaystyle 2\int\frac{u^2}{u^2-2^2}\,dx=2\int\frac{(u^2-2^2)+2^2}{u^2-2^2}\,dx=2\int 1+4\frac{1}{u^2-2^2}\,dx\)

From here, I would consider partial fraction decomposition using the Heaviside cover-up method for simplicity.

Thank you very much!!! That helped me a lot!

 

[pinkcrushin22]

helpful

Hello, pinkcrushin22!

Let \(\displaystyle u \,=\,\sqrt{x}\quad\Rightarrow\quad x \,=\,u^2 \quad\Rightarrow\quad dx \,=\,2u\,du\)

Substitute: .\(\displaystyle \displaystyle \int \frac{u}{u^2-4}(2u\,du) \;=\;2\int\frac{u^2}{u^2-4}\,du \;=\; 2\int\left(1 + \frac{4}{u^2-4}\right)\,du \)

. . . . \(\displaystyle \displaystyle =\; 2\int du + 8 \int\frac{du}{u^2-4} \;=\;2u + \tfrac{8}{4}\ln\left|\frac{u-2}{u+2}\right| + C\)

Back-substitute: .\(\displaystyle 2\sqrt{x} + 2\ln\left|\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\right| + C\)

Thank you so much this was really helpful too.