Johnston Bush
08/22/2024 · Primary School
2. \( \int\left(2 \sqrt{t}-t-\frac{9}{t^{2}}\right) d t \)
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Step-by-step Solution
Calculate the integral \( \int\left(2 \sqrt{t}-t-\frac{9}{t^{2}}\right) dt \).
Evaluate the integral by following steps:
- step0: Evaluate:
\(\int 2\sqrt{t}-t-\frac{9}{t^{2}} dt\)
- step1: Evaluate the power:
\(\int 2t^{\frac{1}{2}}-t-\frac{9}{t^{2}} dt\)
- step2: Use properties of integrals:
\(\int 2t^{\frac{1}{2}} dt-\int t dt-\int \frac{9}{t^{2}} dt\)
- step3: Evaluate the integral:
\(\frac{4}{3}t^{\frac{3}{2}}-\frac{1}{2}t^{2}-\left(-\frac{9}{t}\right)\)
- step4: Evaluate the integral:
\(\frac{4}{3}t^{\frac{3}{2}}-\frac{1}{2}t^{2}+\frac{9}{t}\)
- step5: Simplify:
\(\frac{4}{3}t\sqrt{t}-\frac{1}{2}t^{2}+\frac{9}{t}\)
- step6: Add the constant of integral C:
\(\frac{4}{3}t\sqrt{t}-\frac{1}{2}t^{2}+\frac{9}{t} + C, C \in \mathbb{R}\)
The integral of \( \int\left(2 \sqrt{t}-t-\frac{9}{t^{2}}\right) dt \) is equal to \( \frac{4}{3}t\sqrt{t}-\frac{1}{2}t^{2}+\frac{9}{t} + C \), where \( C \) is an arbitrary constant.
Quick Answer
La integral de \( \int\left(2 \sqrt{t}-t-\frac{9}{t^{2}}\right) dt \) es \( \frac{4}{3}t\sqrt{t}-\frac{1}{2}t^{2}+\frac{9}{t} + C \).
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