1
$\begingroup$

I need to solve

$$\int\frac{2+\sqrt{x}}{x+1}dx$$

I'm forced to use the substitution variable

$$u=\sqrt{x}$$

The first replacement I've done is

$$\int\frac{2+\sqrt{x}}{u^2+1}dx$$

Now, I need to find a way to fit du which is $$du = \frac{1}{2\sqrt{x}}$$ or $$du = \frac{x^\frac{-1}{2}}{2}$$

But I am unable to figure out how to transform the actual $\sqrt{x}$ to the corresponding du.

Any hints ?

$\endgroup$
  • 3
    $\begingroup$ Hint: $u=\sqrt{x}\implies u^2=x\implies 2u\,du=\,dx$ $\endgroup$ – Prasun Biswas Mar 15 '15 at 20:42
  • 1
    $\begingroup$ It seems much more natural to split up the integrand into $$\frac{2}{x+1}+\frac{\sqrt{x}}{x+1}$$ and then use the substitution only on the second part. Is this allowed? $\endgroup$ – Peter Woolfitt Mar 15 '15 at 20:44
  • $\begingroup$ Not sure if it is allowed, but it clearly is easier that way. $\endgroup$ – student Mar 15 '15 at 20:46
1
$\begingroup$

$du/dx = 1/2x^{-1/2} = 1/2u^{-1}$, or equivalently $2u du = dx$, then you get $\int \frac{2+u}{u^2+1}2u du = \int \frac{4u+2u^2}{u^2+1} du$ and then split into two integrals (why are we allowed to do this?):

$$2\int \frac{u^2}{u^2+1}du + 2\int \frac{2u}{1+u^2}du$$

Now you can probably take it from there.

$\endgroup$
1
$\begingroup$

$$u=\sqrt{x}$$ $$x=u^2$$ $$dx=2udu$$ $$\int \frac{2+u}{1+u^2}2udu=\int \frac{4u+2u^2}{1+u^2}du$$ $$\int (\frac{4u}{1+u^2}+\frac{2u^2}{1+u^2})du$$ $$\int (\frac{4u}{1+u^2}+2-\frac{1}{1+u^2})du$$ $$\int (\frac{4u}{1+u^2}+2-\frac{1}{1+u^2})du=2\log(1+u^2)+2u-\tan^{-1}u+C$$ $$=2\log(1+x)+2\sqrt{x}-\tan^{-1}\sqrt{x}+C$$

$\endgroup$
1
$\begingroup$

If $\sqrt{x}=u$ then $x=u^2$ and $dx = 2u\,du$, hence: $$ I=\int \frac{2+\sqrt{x}}{x+1}\,dx = \int \frac{2+u}{1+u^2}\cdot 2u\,du = \int\left(2+\frac{4u-2}{1+u^2}\right)\,du$$ and: $$ I = C+2u+2\log(1+u^2)-2\arctan u = C+2\sqrt{x}+2\log(1+x)-2\arctan\sqrt{x}.$$

$\endgroup$
  • 2
    $\begingroup$ You missed the constant of integration $C$. $\endgroup$ – Prasun Biswas Mar 15 '15 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.