Question

. ∫ d x x + 1 + x ∫ x+1 + x dx

Answer 1

ln|x+√x+1| - (2/√3) arctan((2√x+1)/√3) + (2/3)(x+1)^(3/2) + (2/3)x^(3/2) + C

Answer 2

The given question is ambiguous due to unusual notation. We will interpret the question in the most plausible way that leads to a standard integration problem, typical for JEE/NEET level.

The expression . ∫ d x x + 1 + x ∫ x+1 + x dx is interpreted as the sum of two integrals:

1. The first integral: $\int \frac{dx}{x+1+\sqrt{x}}$ (assuming x after x+1+ refers to $\sqrt{x}$ as is common in such problems where x is also present in the denominator).
2. The second integral: $\int (\sqrt{x+1} + \sqrt{x}) dx$ (assuming x+1 and x inside the second integral refer to $\sqrt{x+1}$ and $\sqrt{x}$ respectively, and the ∫ indicates a separate integral).

Let's solve each integral separately.

Part 1: Solving $I_1 = \int \frac{dx}{x+1+\sqrt{x}}$ Let $u = \sqrt{x}$. Then $u^2 = x$, and $dx = 2u \,du$. Substitute these into the integral: $I_1 = \int \frac{2u \,du}{u^2+1+u} = \int \frac{2u}{u^2+u+1} \,du$ To integrate this, we manipulate the numerator to be the derivative of the denominator ($u^2+u+1$), which is $2u+1$. $I_1 = \int \frac{(2u+1)-1}{u^2+u+1} \,du = \int \frac{2u+1}{u^2+u+1} \,du - \int \frac{1}{u^2+u+1} \,du$

The first part of the integral is a standard logarithmic integral: $\int \frac{2u+1}{u^2+u+1} \,du = \ln|u^2+u+1|$

For the second part, we complete the square in the denominator: $u^2+u+1 = \left(u+\frac{1}{2}\right)^2 + 1 - \left(\frac{1}{2}\right)^2 = \left(u+\frac{1}{2}\right)^2 + \frac{3}{4} = \left(u+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$ This is of the form $\int \frac{1}{y^2+a^2} dy = \frac{1}{a} \arctan\left(\frac{y}{a}\right)$. Here, $y = u+\frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$. So, $\int \frac{1}{u^2+u+1} \,du = \frac{1}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \arctan\left(\frac{2u+1}{\sqrt{3}}\right)$

Combining these, we get: $I_1 = \ln|u^2+u+1| - \frac{2}{\sqrt{3}} \arctan\left(\frac{2u+1}{\sqrt{3}}\right) + C_1$ Substitute back $u=\sqrt{x}$: $I_1 = \ln|x+\sqrt{x}+1| - \frac{2}{\sqrt{3}} \arctan\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right) + C_1$

Part 2: Solving $I_2 = \int (\sqrt{x+1} + \sqrt{x}) dx$ This integral can be solved using the power rule $\int y^n dy = \frac{y^{n+1}}{n+1}$: $I_2 = \int (x+1)^{1/2} dx + \int x^{1/2} dx$ $I_2 = \frac{(x+1)^{1/2+1}}{1/2+1} + \frac{x^{1/2+1}}{1/2+1} + C_2$ $I_2 = \frac{(x+1)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} + C_2$ $I_2 = \frac{2}{3}(x+1)^{3/2} + \frac{2}{3}x^{3/2} + C_2$

Combining the results: The total integral $I = I_1 + I_2$: $I = \ln|x+\sqrt{x}+1| - \frac{2}{\sqrt{3}} \arctan\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right) + \frac{2}{3}(x+1)^{3/2} + \frac{2}{3}x^{3/2} + C$

The final answer is $\boxed{\ln|x+\sqrt{x}+1| - \frac{2}{\sqrt{3}} \arctan\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right) + \frac{2}{3}(x+1)^{3/2} + \frac{2}{3}x^{3/2} + C}$.

Explanation of the solution: The integral is interpreted as the sum of two separate integrals.

1. The first integral $\int \frac{dx}{x+1+\sqrt{x}}$ is solved by substituting $\sqrt{x}=u$, which transforms it into a rational function of $u$. The integrand $\frac{2u}{u^2+u+1}$ is then split into two parts: one whose numerator is the derivative of the denominator (leading to a logarithm), and another which is a standard integral of the form $\frac{1}{y^2+a^2}$ (leading to an arctangent function) after completing the square in the denominator.
2. The second integral $\int (\sqrt{x+1} + \sqrt{x}) dx$ is solved using the simple power rule for integration, treating $(x+1)^{1/2}$ and $x^{1/2}$ as separate terms. The final result is the sum of the results from these two parts.