સંકલન ∫ dx / (1 + √x) શોધો.
Understand the Problem
આ પ્રશ્ન આપણને સંકલન ∫ dx / (1 + √x) શોધવાનું કહે છે. આ સમસ્યાને ઉકેલવા માટે, આપણે યોગ્ય અવેજીકરણનો ઉપયોગ કરીશું અને પછી એકીકરણ કરીશું.
Answer
$2\sqrt{x} - 2\ln(1+\sqrt{x}) + C$
Answer for screen readers
$2\sqrt{x} - 2\ln|1+\sqrt{x}| + C$
Steps to Solve
- Identify the problem and decide on substitution
Integral to be evaluated: $\int \frac{dx}{1+\sqrt{x}}$. Let's use substitution where $u = \sqrt{x}$, which means $x = u^2$.
- Compute the differential
Compute $dx$ in terms of $du$: $x = u^2$ Differentiating both sides wrt $u$, we get $\frac{dx}{du} = 2u$ Therefore, $dx = 2u , du$
- Substitute into the integral
Now we substitute $u$ and $du$ in the integral: $\int \frac{dx}{1+\sqrt{x}} = \int \frac{2u , du}{1+u}$
- Rewrite the integral to simplify
Rewrite $2u$ as $2(u+1-1) = 2(u+1) - 2$. $\int \frac{2u , du}{1+u} = \int \frac{2(u+1) - 2}{1+u} , du = \int \left(2 - \frac{2}{1+u}\right) , du$
- Split and integrate
Now integrate term by term: $\int \left(2 - \frac{2}{1+u}\right) , du = \int 2 , du - \int \frac{2}{1+u} , du = 2\int du - 2\int \frac{1}{1+u} , du$ $= 2u - 2\ln|1+u| + C$
- Back-substitute for final answer
Finally, substitute back $u = \sqrt{x}$: $2u - 2\ln|1+u| + C = 2\sqrt{x} - 2\ln|1+\sqrt{x}| + C$
$2\sqrt{x} - 2\ln|1+\sqrt{x}| + C$
More Information
The integral $\int \frac{dx}{1+\sqrt{x}}$ simplifies to $2\sqrt{x} - 2\ln|1+\sqrt{x}| + C$, where $C$ is the constant of integration. Note that since $\sqrt{x}$ is always non-negative, we can write $2\sqrt{x} - 2\ln(1+\sqrt{x}) + C$, omitting the absolute value.
Tips
A common mistake is forgetting to substitute back to the original variable $x$ after integrating with respect to $u$. Also, one might forget the constant of integration $C$. Another mistake involves the algebraic manipulation to simplify the integral before integrating.
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