5.6 Integrals involving exponential and logarithmic functions  (Page 3/4)

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x , then bring the x ² in the denominator up to the numerator using a negative exponent. We have

Let $u=x^{\mathrm{-1}},$ the exponent on e . Then

Bringing the negative sign outside the integral sign, the problem now reads

Next, change the limits of integration:

Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,

First factor the 3 outside the integral symbol. Then use the u ⁻¹ rule. Thus,

See [link] .

This can be rewritten as ${\displaystyle \int \left(2x^3+3x\right){\left(x^4+3x^2\right)}^{\mathrm{-1}}dx}.$ Use substitution. Let $u=x^4+3x^2,$ then $du=4x^3+6x.$ Alter du by factoring out the 2. Thus,

Rewrite the integrand in u :

Then we have

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

We need substitution to evaluate this problem. Let $u=1+\text{cos}x,,$ so $du=\text{−}\text{sin}xdx.$ Rewrite the integral in terms of u , changing the limits of integration as well. Thus,

Then