`int(x+1)sqrt(2x-1)dx=`

Options

A. `(1)/(10)(2x-1)^(3//2)+(1)/(2)(2x-1)^(5//2)+c`
B. `(1)/(10)(2x-1)^(5//2)-(1)/(2)(2x-1)^(3//2)+c`
C. `(1)/(10)(x+1)^(5//2)+(1)/(2)(x+1)^(3//2)+c`
D. `(1)/(10)(x+1)^(5//2)+(1)/(2)(x+1)^(3//2)+c`

Answer

To solve the integral \( \int (x + 1) \sqrt{2x - 1} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( y^2 = 2x - 1 \). Then, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = 2 \implies 2y \frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y} \]
Thus, we have:
\[ dx = y \, dy \]
Now, we can express \( x \) in terms of \( y \):
\[ 2x = y^2 + 1 \implies x = \frac{y^2 + 1}{2} \]

### Step 2: Substitute in the Integral
Now we substitute \( x \) and \( dx \) into the integral:
\[ \int (x + 1) \sqrt{2x - 1} \, dx = \int \left( \frac{y^2 + 1}{2} + 1 \right) y \cdot y \, dy \]
This simplifies to:
\[ = \int \left( \frac{y^2 + 1 + 2}{2} \right) y^2 \, dy = \int \left( \frac{y^2 + 3}{2} \right) y^2 \, dy \]
\[ = \frac{1}{2} \int (y^4 + 3y^2) \, dy \]

### Step 3: Integrate
Now we integrate:
\[ = \frac{1}{2} \left( \frac{y^5}{5} + \frac{3y^3}{3} \right) + C \]
\[ = \frac{1}{2} \left( \frac{y^5}{5} + y^3 \right) + C \]
\[ = \frac{y^5}{10} + \frac{y^3}{2} + C \]

### Step 4: Substitute Back
Now we substitute back \( y = \sqrt{2x - 1} \):
\[ = \frac{(\sqrt{2x - 1})^5}{10} + \frac{(\sqrt{2x - 1})^3}{2} + C \]
\[ = \frac{(2x - 1)^{5/2}}{10} + \frac{(2x - 1)^{3/2}}{2} + C \]

### Final Answer
Thus, the final result is:
\[ \int (x + 1) \sqrt{2x - 1} \, dx = \frac{(2x - 1)^{5/2}}{10} + \frac{(2x - 1)^{3/2}}{2} + C \]

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