[tex]\int \frac{y^2}{2\sqrt{y^3+5}} \, dx[/tex]
[tex]\int \frac{5}{x} \cot(\ln x^3) \, dx[/tex]
[tex]\int \frac{e^{x^2-2x}(2x-5)}{e^{1+3x}} \, dx[/tex]
[tex]\int \frac{dz}{z^2+6z+5}[/tex]
[tex]\int \frac{dx}{3+2^x}[/tex]
[tex]\int \frac{5dw}{w\sqrt{9w^6-4}}[/tex]
Answer (1)
To solve the integrals given in the question, it is important to address each one separately. Here is a step-by-step breakdown for each integral:
∫ 2 y 3 + 5 y 2 d x
This integral involves calculus techniques and appears to need a substitution to simplify the expression under the square root. Let u = y 3 + 5 . Then, d u = 3 y 2 d y . This substitution simplifies the integral to a form that can be more easily solved.
∫ x 5 cot ( ln x 3 ) d x
The cotangent function inside the integral suggests possible trigonometric identities that might simplify the problem. Recognizing the properties of logarithms and trigonometric functions often helps in these cases.
∫ e 1 + 3 x e x 2 − 2 x ( 2 x − 5 ) d x
The exponential functions in the numerator and denominator should be simplified using exponent rules like e a / e b = e a − b . This approach often makes integration more straightforward.
∫ z 2 + 6 z + 5 d z
Factoring the quadratic expression in the denominator will help convert the integral into partial fractions, which is a standard technique to solve rational function integrals.
∫ 3 + 2 x d x
This integral may require substitution like u = 2 x to simplify the expression. The logarithmic integral approximation might be necessary if no closed-form solution is directly apparent.
∫ w 9 w 6 − 4 5 d w
Consider substituting v = 9 w 6 − 4 to transform the integrand into a more recognizable form. This change makes it possible to tackle the integral using basic integral rules.
Each integral uses different integration techniques, such as substitution, integration by parts, or using partial fractions, depending on the form and complexity of the function. Solving each requires careful algebra and calculus manipulations.
