(2 sqrt(x)-5e^(-3x)+4secxtanx+3/x​)dx

asked by guest
on Jan 15, 2025 at 8:58 am



You asked:

Evaluate the expression: \(\left(2 \sqrt{x} - 5 {e}^{-\left( 3 x \right)} + 4 \sec\left( x \right) \tan\left( x \right) + \frac{3}{x}\right) \cdot d x\)

MathBot Answer:

Evaluated



\(\displaystyle \left(2 \sqrt{x} - 5 {e}^{-\left( 3 x \right)} + 4 \sec\left( x \right) \tan\left( x \right) + \frac{3}{x}\right) \cdot d x = d x \left(2 \sqrt{x} + 4 \tan{\left(x \right)} \sec{\left(x \right)} - 5 e^{- 3 x} + \frac{3}{x}\right) \)

Expanded

\[\left(2 \sqrt{x} - 5 {e}^{-\left( 3 x \right)} + 4 \sec\left( x \right) \tan\left( x \right) + \frac{3}{x}\right) \cdot d x = 2 x^{\frac{3}{2}} d + 4 d x \tan{\left(x \right)} \sec{\left(x \right)} - 5 d x e^{- 3 x} + 3 d\]

Factored

\[\left(2 \sqrt{x} - 5 {e}^{-\left( 3 x \right)} + 4 \sec\left( x \right) \tan\left( x \right) + \frac{3}{x}\right) \cdot d x = d \left(2 x^{\frac{3}{2}} e^{3 x} + 4 x e^{3 x} \tan{\left(x \right)} \sec{\left(x \right)} - 5 x + 3 e^{3 x}\right) e^{- 3 x}\]

\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).

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