Problem:
Evaluate the following integral:
$$ \int \frac{\sqrt{x}}{1+x} \, dx $$
Answer:
Let $I$ be the integral we are evaluating. We are going to use the substitution $ \tan u = \sqrt{x}$. We have:
\begin{align*}
\sec^2{u} \,\, du &= \left( \frac{1}{2} \right) x^{ - \frac{1}{2} } \, dx \\
I &= 2 \int \frac{ \left( \frac{1}{2} \left( \frac{x}{ \sqrt{x} }\right) \right) }{1+x} \, dx \\
I &= 2 \int \frac{\tan^2{u}} {\tan^2{u}+ 1} \, du = 2 \int \frac{ \tan^2 u } { \sec^2 u } \, du \\
I &= 2 \int \sin^2 u \, du \\
\end{align*}
Now recall the trig idenity:
$$ \sin^2 \theta = \frac{2 - \cos{2 \theta} }{2} $$
\begin{align*}
I &= 2 \int \frac{2 - \cos{2 u} }{2} \, du = \int 2 - \cos{2 u} \, du \\
I &= 2u - \frac{\sin{2u}}{2} + C = 2u - \frac{\sqrt{ 1 - \cos^2{2u}}}{2} + C
\end{align*}
The book's answer is:
$$ \int \frac{\sqrt{x}}{1+x} \, dx = 2\sqrt{x} - 2\tan{( \sqrt{x}) } + C $$
I made a mistake in entering the book's answer. Here is what it should be:
The book's answer is:
$$ \int \frac{\sqrt{x}}{1+x} \, dx = 2\sqrt{x} - 2\tan^{-1}{( \sqrt{x}) } + C $$
Therefore, I conclude my answer is wrong. Where did I go wrong?
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2$\begingroup$ You have to use a trig substitution ? $u=\sqrt x$ is a better substitution $\endgroup$ – Satyendra Mar 8 '20 at 2:22
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$\begingroup$ I do not have to use a trig substitution but when I tired $u = \sqrt{x}$ it did not work out so well. I am going to try again. $\endgroup$ – Bob Mar 8 '20 at 2:25
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2$\begingroup$ What do you mean it didnt work ? $u=\sqrt x \implies du =\frac 12 \frac {dx}{\sqrt x}$ $\endgroup$ – Satyendra Mar 8 '20 at 2:27
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1$\begingroup$ The book's answer is WRONG! "tan" should be "arctan" $\endgroup$ – P. Lawrence Mar 8 '20 at 2:28
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$\begingroup$ The book has the right answer. I typed it in wrong. I will fix the post. $\endgroup$ – Bob Mar 8 '20 at 2:32
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$$I=\int \frac{\sqrt{x}}{1+x} \, dx$$ Substitute $u=\sqrt x \implies dx=2udu $ $$I=2\int \frac{{u^2}}{1+u^2} \, du=2\int \frac{{u^2+1-1}}{1+u^2} \, du$$ $$I =2 \left(u-\int \frac{du}{1+u^2} \right)$$ $$I =2(u-\arctan u) +C =2(\sqrt{x}-\arctan \sqrt{x}) +C$$
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You made a mistake during the substitution. With $\tan u = \sqrt{x}$, you get $x = \tan^2u$ and $dx = 2\tan u \sec^2udu$. Then,
$$ \int \frac{\sqrt{x}}{1+x}dx = \int \frac{\tan^2u \cdot 2\sec^2u}{1+\tan^2u}du = 2\int \tan^2udu = 2\int (\sec^2u-1)du $$
$$=2(\tan u-u)+C = 2(\sqrt x - \tan^{-1}\sqrt x)+C$$
