Question

Use the formula ∫ f⁻¹(x) dx = x f⁻¹(x) - ∫ f(y) dy, y = f⁻¹(x)To evaluate the integrals in Exercises 77-80. Express your answers in terms of x.∫ log₂ x dx

Accepted Answer

Identify the function inside the integral: here, we have $$ \int \log_2 x \, dx . $$Recognize that $$ \log_2 x  is $$the inverse of the exponential function $$ f(y) = 2^y $$, since $$ y = \log_2 x $$ implies $$ x = 2^y . $$Recall the formula for integrating an inverse function: $$ \int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(y) \, dy, \quad 	ext{where } y = f^{-1}(x).  In $$this problem, $$ f^{-1}(x) = \log_2 x $$ and $$ f(y) = 2^y . $$Substitute into the formula: $$ \int \log_2 x \, dx = x \log_2 x - \int 2^y \, dy, \quad 	ext{with } y = \log_2 x. $$ This breaks the original integral into two parts: the product $$ x \log_2 x $$ and the integral of $$ 2^y $$ with respect to $$ y . $$Evaluate the integral $$ \int 2^y \, dy . $$Recall that the integral of an exponential function with base $$ a  is$$ $$ \int a^y \, dy = \frac{a^y}{\ln a} + C . So$$, $$ \int 2^y \, dy = \frac{2^y}{\ln 2} + C. $$ Finally, substitute back $$ y = \log_2 x $$ into the expression to write the answer entirely in terms of $$ x . $$This will give you the integral $$ \int \log_2 x \, dx $$ expressed in terms of $$ x .$$

Use the formula ∫ f⁻¹(x) dx = x f⁻¹(x) - ∫ f(y) dy, y = f⁻¹(x)
To evaluate the integrals in Exercises 77-80. Express your answers in terms of x.
∫ log₂ x dx

Verified video answer for a similar problem:

Key Concepts

Inverse Function Integration Formula

Properties of Logarithmic Functions

Integration by Parts