What is the integral of ((a^x-b^x)^2)/(a^xb^x) ?

The integral of ((a^x-b^x)^2)/(a^xb^x) is (a^x)/(b^xln(a/b))-2x+(b^x)/(a^xln(b/a))+C

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integral of ((a^x-b^x)^2)/(a^xb^x)

\int \frac{( a ^{x} - b ^{x} ) ^{2}}{a ^{x} b ^{x}} d x

\frac{a ^{x}}{b ^{x} ln ( \frac{a}{b} )} - 2 x + \frac{b ^{x}}{a ^{x} ln ( \frac{b}{a} )} + C

Solution steps

\int \frac{( a ^{x} - b ^{x} ) ^{2}}{a ^{x} b ^{x}} d x

Expand

\frac{( a ^{x} - b ^{x} ) ^{2}}{a ^{x} b ^{x}} : \frac{a ^{x}}{b ^{x}} - 2 + \frac{b ^{x}}{a ^{x}}

= \int \frac{a ^{x}}{b ^{x}} - 2 + \frac{b ^{x}}{a ^{x}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{a ^{x}}{b ^{x}} d x - \int 2 d x + \int \frac{b ^{x}}{a ^{x}} d x

\int \frac{a ^{x}}{b ^{x}} d x = \frac{a ^{x}}{b ^{x} ln ( \frac{a}{b} )}

\int 2 d x = 2 x

\int \frac{b ^{x}}{a ^{x}} d x = \frac{b ^{x}}{a ^{x} ln ( \frac{b}{a} )}

= \frac{a ^{x}}{b ^{x} ln ( \frac{a}{b} )} - 2 x + \frac{b ^{x}}{a ^{x} ln ( \frac{b}{a} )}

Add a constant to the solution

= \frac{a ^{x}}{b ^{x} ln ( \frac{a}{b} )} - 2 x + \frac{b ^{x}}{a ^{x} ln ( \frac{b}{a} )} + C

\frac{d y}{d x} = e^{7 x - 7 y}

\frac{\partial}{\partial z} (2 z)

y^{^{'}} + y = 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 - t

\int \frac{x}{x ^{3}} d x

What is the integral of ((a^x-b^x)^2)/(a^xb^x) ?
The integral of ((a^x-b^x)^2)/(a^xb^x) is (a^x)/(b^xln(a/b))-2x+(b^x)/(a^xln(b/a))+C