# How to evaluate $\int \frac{1+x}{1+\sqrt x} dx$

I picked this exercice from B. Demidovitch and i started solving, but when it's complete, i get only the half of the solution (the first two fractions and the ln, doesn't show up). $$\int \frac{1+x}{1+\sqrt x} dx$$

The Solution is:

$$2 \left[\frac{\sqrt{x^3}}{3}-\frac{x}{2}+2\sqrt{x}-2ln(1+\sqrt{x})\right] + C$$

• Hint: Partial Fractions. – Nobody Aug 15 '16 at 16:05
• Try the supstitution $t = \sqrt{x}$. – Icarus 369 Aug 15 '16 at 16:06
• @Icarus369 already did -, that was the first thing i tryed. – Edilson Aug 15 '16 at 16:08
• @Edilson You're on the right track then. Since $\frac{1+x}{1+\sqrt{x}}$ can be expanded as $\sqrt{x}+\frac{2}{\sqrt{x}+1}-1$ it seems you made an error in your original attempt. Try carrying out the rest of the nice answer below. You'll have your antiderivative when you can differentiate it and obtain $\frac{1+x}{1+\sqrt{x}}$. – Nobody Aug 15 '16 at 16:13
• Now i'm doing $t=\sqrt{x} + 1$, like in the answer below. – Edilson Aug 15 '16 at 16:20

Let $u=\sqrt x+1$, so that $\mathrm{d}x=2(u-1)\,\mathrm{d}u$. $$\int\frac{1+x}{1+\sqrt x}\,\mathrm{d}x=\int\frac{1+(u-1)^2}{u}(2(u-1))\,\mathrm{d}u$$ Expanding makes the integral trivial. (added some more details below) \begin{align*}2\int\frac{u-1+(u-1)^3}{u}\,\mathrm{d}u&=2\int\frac{u^3-3u^2+4u-2}{u}\,\mathrm{d}u\\[1ex]&=2\int\left(u^2-3u+4-\frac{2}{u}\right)\,\mathrm{d}u\\[1ex]&=\frac{2}{3}u^3-3u^2+8u-4\ln|u|+C\\[1ex]&=\frac{2}{3}\left(\sqrt x+1\right)^3-3\left(\sqrt x+1\right)^2+8\left(\sqrt x+1\right)-4\ln\left|\sqrt x+1\right|+C\\[1ex]&=\frac{2}{3}x^{3/2}-x+4x^{1/2}-4\ln\left(\sqrt x+1\right)+C\end{align*}

Instead of @user170231 (which is a sutable substitution) I should use (as you tried also):

$$u=\sqrt{x}\space\space\space\space\space\space\space\text{and}\space\space\space\space\space\space\space\text{d}u=\frac{1}{2\sqrt{x}}\space\text{d}x$$

So, we get:

$$\text{I}=\int\frac{1+x}{1+\sqrt{x}}\space\text{d}x=2\int\frac{u(u^2+1)}{u+1}\space\text{d}u$$

With long division we find:

$$\frac{u(u^2+1)}{u+1}=2+u^2-u-\frac{2}{1+u}$$

Now, we get:

$$\text{I}=2\left[2\int1\space\text{d}u+\int u^2\space\text{d}u-\int u\space\text{d}u-2\int\frac{1}{1+u}\space\text{d}u\right]$$

Use:

• $$\int1\space\text{d}u=u+\text{C}$$
• $$\int u^2\space\text{d}u=\frac{u^3}{3}+\text{C}$$
• $$\int u\space\text{d}u=\frac{u^2}{2}+\text{C}$$
• For $\int\frac{1}{1+u}\space\text{d}u$ substitute $s=1+u$ and $\text{d}s=\text{d}u$: $$\int\frac{1}{1+u}\space\text{d}u=\int\frac{1}{s}\space\text{d}s=\ln\left|s\right|+\text{C}$$

So:

$$\text{I}=2\left[2\sqrt{x}+\frac{u^3}{3}-\frac{u^2}{2}-2\ln\left|s\right|\right]+\text{C}=2\left[2\sqrt{x}+\frac{\left(\sqrt{x}\right)^3}{3}-\frac{\left(\sqrt{x}\right)^2}{2}-2\ln\left|1+\sqrt{x}\right|\right]+\text{C}$$

By setting $\color{red}{x=z^2}$ we just have to compute

$$I=\int \frac{2z(1+z^2)}{1+z}\,dz$$ and by polynomial division we have $2z(1+z^2)=(2z^2-2z+4)(1+z)-4$, hence $$I = \int (2z^2-2z+4)\,dz -4\int\frac{dz}{1+z} = \color{red}{\frac{z^3}{3}-z^2+4-4\log(1+z)+C}.$$