How do I apply $u$-substitution to solve $\int x \ln\left(1+\sqrt{x}\right) \,dx$

Question

$$\int x \ln\left(1+\sqrt{x}\right) \,dx$$

I think u-substitution would be the best method but I'm not entirely sure how I would go about using it.

Answer 1

HINT:

$$\int x\ln\left(1+\sqrt{x}\right)\space\text{d}x=$$

---

Use integration by parts, $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where:

$$f=\ln\left(1+\sqrt{x}\right),\text{d}g=x\space\text{d}x,\text{d}f=\frac{1}{2\left(x+\sqrt{x}\right)}\space\text{d}x,g=\frac{x^2}{2}$$

---

$$\frac{x^2\ln\left(1+\sqrt{x}\right)}{2}-\frac{1}{2}\int\frac{x^{\frac{3}{2}}}{2\sqrt{x}+2}\space\text{d}x=$$

---

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

---

$$\frac{x^2\ln\left(1+\sqrt{x}\right)}{2}-\int\frac{u^4}{2u+2}\space\text{d}u$$

Now, use long division.

Answer 2

If we set $u = 1 + \sqrt{x}$, we can rearrange to get $x = (u - 1)^2$, so $dx = 2 (u - 1) \,du$, giving $$\int (u - 1)^2 \log u \cdot 2 (u - 1) \,du = 2 \int (u - 1)^3 \log u \,du .$$ This is a product of two types of functions and so suggests using integration by parts; making the only sensible choices for $v$ and $dw$ yields an integral of a sum of power functions.

Answer 3

Apply Integral Substitution: $\color{green}{u=\sqrt{x}\quad \:du=\frac{1}{2u}dx}$

$$\int \:x\ln \left(1+\sqrt{x}\right)dx=2\int \:u^3\ln \left(u+1\right)du$$

Now you can integrate by parts...