Q.Evaluate: $\int \frac{1 + x}{x} + e^{-x}\,dx$

Expand $\left(x+\frac{1}{x}\right)^3$ using $(a+b)^3$, then integrate each power of $x$ separately.

First expand the cube using $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ with $a=x$, $b=\frac{1}{x}$:

$\left(x + \frac{1}{x}\right)^3 = x^3 + 3x^2\cdot\frac{1}{x} + 3x\cdot\frac{1}{x^2} + \frac{1}{x^3} = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}.$

Now integrate term by term:

$\int x^3\,dx = \frac{x^4}{4}, \quad \int 3x\,dx = \frac{3x^2}{2}, \quad \int \frac{3}{x}\,dx = 3\log x,$

$\int \frac{1}{x^3}\,dx = \int x^{-3}\,dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}.$

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Differentiating $Ax - 3\log|2e^x - 5| + c$ and equating it to the integrand $\dfrac{4e^x - 25}{2e^x - 5}$ forces $A = 5$, so the statement is True.

The quickest check is to differentiate the proposed answer and compare with the integrand. Differentiating $F(x) = Ax - 3\log|2e^x - 5| + c$:

$F'(x) = A - 3\cdot\frac{2e^x}{2e^x - 5} = A - \frac{6e^x}{2e^x - 5}.$

This must equal the integrand $\dfrac{4e^x - 25}{2e^x - 5}$. Putting the left side over the common denominator:

$\frac{A(2e^x - 5) - 6e^x}{2e^x - 5} = \frac{4e^x - 25}{2e^x - 5}.$

Equating numerators: …