Final Answer
$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$
Got another answer? Verify it here!
Step-by-step Solution
Specify the solving method
Choose an option Integrate using basic integrals Integrate by substitution Integrate by parts Tabular Integration Suggest another method or feature
Send
1
We can solve the integral $\int\ln\left(\sqrt{x}+\sqrt{1+x}\right)dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
Intermediate steps
2
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=\ln\left(\sqrt{x}+\sqrt{1+x}\right)}\\ \displaystyle{du=\frac{1}{2\sqrt{x}\sqrt{1+x}}dx}\end{matrix}$
Explain this step further
3
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=1dx}\\ \displaystyle{\int dv=\int 1dx}\end{matrix}$
5
The integral of a constant is equal to the constant times the integral's variable
$x$
Intermediate steps
6
Now replace the values of $u$, $du$ and $v$ in the last formula
$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)-\int\frac{\sqrt{x}}{2\sqrt{1+x}}dx$
Explain this step further
Intermediate steps
7
The integral $-\int\frac{\sqrt{x}}{2\sqrt{1+x}}dx$ results in: $-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)$
$-\frac{1}{2}\sqrt{x}\sqrt{1+x}+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)$
Explain this step further
8
Gather the results of all integrals
$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}$
9
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$
Final Answer$x\ln\left(\sqrt{x}+\sqrt{1+x}\right)+\frac{1}{2}\ln\left(\sqrt{1+x}+\sqrt{x}\right)-\frac{1}{2}\sqrt{x}\sqrt{1+x}+C_0$