Question

Find each indefinite integral

a. ∫ e^x(e^(3x)-5) dx

b. ∫ x(sqrt(2x^2+3) dx

c. ∫ (x)/(sqrt(3-x)) dx

Answer 1

Answer on Question #48163 – Math – Integral Calculus

a) $\int e^{x}\left(e^{3x} - 5\right)dx.$

**Solution.**

$$\int e^{x}\left(e^{3x} - 5\right)dx = \int \left(e^{x} \cdot e^{3x} - e^{x} \cdot 5\right)dx = \int \left(e^{4x} - 5e^{x}\right)dx = \int e^{4x}dx - 5\int e^{x}dx.$$ $$\int e^{4x}dx = \left[ \begin{array}{l} t = 4x \\ dt = 4dx \Rightarrow dx = \frac{dt}{4} \end{array} \right] = \int \frac{e^{t}}{4}dt = \frac{e^{t}}{4} + C = \frac{1}{4}e^{4x} + C;$$ $$\int e^{x}dx = e^{x} + C$$

Hence,

$$\int e^{x}\left(e^{3x} - 5\right)dx = \frac{1}{4}e^{4x} - 5e^{x} + C.$$

**Answer:** $\frac{1}{4}e^{4x} - 5e^{x} + C.$

b) $\int x\left(\sqrt{2x^2 + 3}\right)dx.$

**Solution.**

$$\begin{aligned} \int x\left(\sqrt{2x^2 + 3}\right)dx &= \left[ \begin{array}{l} t = 2x^2 + 3 \\ dt = \left(2x^2 + 3\right)'dx = 4xdx \\ xdx = \frac{1}{4}dt \end{array} \right] = \int \frac{1}{4}\sqrt{t}dt = \frac{1}{4}\int t^{\frac{1}{2}}dt = \\ &= \frac{1}{4} \cdot \frac{t^{\frac{1}{2} + 1}}{2} + C = \frac{1}{4} \cdot \frac{t^{\frac{3}{2}}}{2} + C = \frac{2}{4} \cdot \frac{(2x^2 + 3)^{\frac{3}{2}}}{2} + C = \frac{(2x^2 + 3)^{\frac{3}{2}}}{6} + C. \\ &\quad \frac{(2x^2 + 3)^{\frac{3}{2}}}{6} + C. \end{aligned}$$

c) $\int \frac{x}{\sqrt{3 - x}} dx.$

**Solution.**

$$\int \frac {x}{\sqrt {3 - x}} d x = \int \frac {x - 3 + 3}{\sqrt {3 - x}} d x = \int \frac {- (3 - x)}{\sqrt {3 - x}} d x + \int \frac {3}{\sqrt {3 - x}} d x = - \int \sqrt {3 - x} d x + 3 \int \frac {d x}{\sqrt {3 - x}}.$$ $$\int \sqrt {3 - x} d x = \left[ \begin{array}{l} t = 3 - x \\ d t = - d x \end{array} \right] = - \int \sqrt {t} d t = - \frac {t ^ {\frac {1}{2} + 1}}{\frac {1}{2} + 1} + C = - \frac {t ^ {\frac {3}{2}}}{\frac {3}{2}} + C = - \frac {2 (3 - x) ^ {\frac {3}{2}}}{3} + C;$$ $$\int \frac {d x}{\sqrt {3 - x}} = \left[ \begin{array}{l} t = 3 - x \\ d t = - d x \end{array} \right] = - \int \frac {d t}{\sqrt {t}} = - \frac {t ^ {\frac {1}{2} + 1}}{- \frac {1}{2} + 1} + C = - \frac {t ^ {\frac {1}{2}}}{\frac {1}{2}} + C = - 2 \sqrt {3 - x} + C$$

Hence,

$$\int \frac {x}{\sqrt {3 - x}} d x = \frac {2 (3 - x) ^ {\frac {3}{2}}}{3} + 3 (- 2 \sqrt {3 - x} + C) = \frac {2}{3} \sqrt {(3 - x) ^ {3}} - 6 \sqrt {3 - x} + C$$

Answer: $\frac{2}{3}\sqrt{(3 - x)^3} - 6\sqrt{3 - x} + C$

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