Define Integral: Let's denote the integral we want to evaluate as I:I=∫−6x4−4xdxTo solve this integral, we will use integration by parts, which states that ∫udv=uv−∫vdu, where u and dv are differentiable functions of x. We need to choose u and dv such that the resulting integral is easier to solve. Let's choose:u=−6x (which means du=−6dx)dv=4−4xdx (which means we need to find ∫udv=uv−∫vdu0)To find ∫udv=uv−∫vdu0, we need to integrate dv. Since ∫udv=uv−∫vdu3 is an exponential function, we can use the fact that ∫udv=uv−∫vdu4 when ∫udv=uv−∫vdu5 and ∫udv=uv−∫vdu6. In our case, ∫udv=uv−∫vdu7 and ∫udv=uv−∫vdu8, so ∫udv=uv−∫vdu9.
Integration by Parts: Now we integrate dv to find v: v=∫4(−4x)dx Using the formula for integrating an exponential function with a linear exponent, we get: v=(−4ln(4))4(−4x) We can simplify this by multiplying the numerator and denominator by −1: v=−4ln(4)4(−4x)
Choose u and dv: Now that we have u and v, we can apply the integration by parts formula:I=uv−∫vduSubstituting u, du, and v into the formula, we get:I=−6x(−4−4x/(4ln(4)))−∫(−4−4x/(4ln(4)))(−6dx)Simplifying the integral, we get:I=4ln(4)6x4−4x+4ln(4)6∫4−4xdx
Find v: We have already integrated 4−4x when finding v, so we can use that result here:I=4ln(4)6x4−4x+4ln(4)6(−4ln(4)4−4x)Simplifying the expression, we get:I=2ln(4)3x4−4x−8ln(4)23⋅4−4x
Apply Integration by Parts: Now we can combine the terms with a common base of 4−4x:I=2ln(4)3x4−4x−8ln(4)23⋅4−4x$I = \(4\)^{\(-4\)x}\left(\frac{\(3\)x}{\(2\)\ln(\(4\))} - \frac{\(3\)}{\(8\)\ln(\(4\))^\(2\)}\right)
Simplify Integral: Finally, we add the constant of integration \(C\) to our result:\(\newline\)\(I = 4^{-4x}\left(\frac{3x}{2\ln(4)} - \frac{3}{8\ln(4)^2}\right) + C\)\(\newline\)This is the indefinite integral of the given function.
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