Carrillo Bond
09/19/2023 · Primary School
integrales teniendo en cuenta la forma \( \sqrt{u^{2} \pm a^{2}}, \sqrt{a^{2}-u^{2}}, u^{2} \pm a^{2}, a^{2}-u^{2} \) \[ \begin{array}{ll}\text { Cristian } & \text { [6) } \int \frac{d x}{9 x^{2}-144} \\ \text { (arina } & \text { 7) } \int \frac{d x}{\sqrt{25-9 x^{2}}} \\ \text { 8) } & \int \frac{d x}{\sqrt{4 x^{2}-7}}\end{array} \]
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Para resolver las integrales dadas, utilizaremos las formas \( \sqrt{u^{2} \pm a^{2}}, \sqrt{a^{2}-u^{2}}, u^{2} \pm a^{2}, a^{2}-u^{2} \).
1. Para la integral \( \int \frac{d x}{9 x^{2}-144} \):
- Identificamos que \( u = 3x \) y \( a = 12 \).
- La integral se convierte en \( \int \frac{d u}{u^{2}-144} \).
- Utilizamos la forma \( \frac{1}{a} \ln \left| \frac{u-a}{u+a} \right| \).
- La integral se convierte en \( \frac{1}{24} \ln \left| \frac{3x-12}{3x+12} \right| + C \).
2. Para la integral \( \int \frac{d x}{\sqrt{25-9 x^{2}}} \):
- Identificamos que \( u = 3x \) y \( a = 5 \).
- La integral se convierte en \( \int \frac{d u}{\sqrt{a^{2}-u^{2}}} \).
- Utilizamos la forma \( \frac{1}{a} \arcsin \left( \frac{u}{a} \right) \).
- La integral se convierte en \( \frac{1}{5} \arcsin \left( \frac{3x}{5} \right) + C \).
3. Para la integral \( \int \frac{d x}{\sqrt{4 x^{2}-7}} \):
- Identificamos que \( u = 2x \) y \( a = \sqrt{7} \).
- La integral se convierte en \( \int \frac{d u}{\sqrt{a^{2}-u^{2}}} \).
- Utilizamos la forma \( \frac{1}{a} \arcsin \left( \frac{u}{a} \right) \).
- La integral se convierte en \( \frac{1}{2\sqrt{7}} \arcsin \left( \frac{2x}{\sqrt{7}} \right) + C \).
Por lo tanto, las soluciones de las integrales son:
1. \( \frac{1}{24} \ln \left| \frac{3x-12}{3x+12} \right| + C \)
2. \( \frac{1}{5} \arcsin \left( \frac{3x}{5} \right) + C \)
3. \( \frac{1}{2\sqrt{7}} \arcsin \left( \frac{2x}{\sqrt{7}} \right) + C \)
Quick Answer
1. \( \frac{1}{24} \ln \left| \frac{3x-12}{3x+12} \right| + C \)
2. \( \frac{1}{5} \arcsin \left( \frac{3x}{5} \right) + C \)
3. \( \frac{1}{2\sqrt{7}} \arcsin \left( \frac{2x}{\sqrt{7}} \right) + C \)
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