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
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
We can solve the integral $\int\frac{\sqrt{x}}{2\sqrt{1+x}}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2\sqrt{1+x}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
$u=2\sqrt{1+x}$
Intermediate steps
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Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above
$du=\left(1+x\right)^{-\frac{1}{2}}dx$
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Isolate $dx$ in the previous equation
$\frac{du}{\left(1+x\right)^{-\frac{1}{2}}}=dx$
Intermediate steps
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Rewriting $x$ in terms of $u$
$x=\frac{u^{2}}{4}-1$
Intermediate steps
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Substituting $u$, $dx$ and $x$ in the integral and simplify
The integral $-\int\frac{\sqrt{u^{2}-4}}{4}du$ results in: $-\frac{1}{4}\sqrt{1+x}\sqrt{4\left(1+x\right)-4}+\frac{1}{2}\ln\left(\sqrt{1+x}+\frac{\sqrt{\left(2\sqrt{1+x}\right)^{2}-4}}{2}\right)$
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more