What is the integral of (e^{2x}(1+2xln(x)))/(4x) ?

The integral of (e^{2x}(1+2xln(x)))/(4x) is 1/4 e^{2x}ln(x)+C

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integral of (e^{2x}(1+2xln(x)))/(4x)

\int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{4 x} d x

\frac{1}{4} e^{2 x} ln (x) + C

Solution steps

\int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{4 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{x} d x

Expand

\frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{x} : \frac{e ^{2 x}}{x} + 2 e^{2 x} ln (x)

= \frac{1}{4} \cdot \int \frac{e ^{2 x}}{x} + 2 e^{2 x} ln (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

= \frac{1}{4} (\int \frac{e ^{2 x}}{x} d x + \int 2 e^{2 x} ln (x) d x)

\int \frac{e ^{2 x}}{x} d x = Ei (2 x)

\int 2 e^{2 x} ln (x) d x = e^{2 x} ln (x) - Ei (2 x)

= \frac{1}{4} (Ei (2 x) + e^{2 x} ln (x) - Ei (2 x))

Simplify

\frac{1}{4} (Ei (2 x) + e^{2 x} ln (x) - Ei (2 x)) : \frac{1}{4} e^{2 x} ln (x)

= \frac{1}{4} e^{2 x} ln (x)

Add a constant to the solution

= \frac{1}{4} e^{2 x} ln (x) + C

\int \frac{x}{( 4 - x ^{2} ) ^{2}} d x

\frac{4 x}{x + 1}

\int_{1}^{1.5} (\frac{1}{8} + \frac{3}{8} x) d x

f (t) = 4 t^{2} - 5 sin (3 t)

x \to - \infty lim (x^{1})

What is the integral of (e^{2x}(1+2xln(x)))/(4x) ?
The integral of (e^{2x}(1+2xln(x)))/(4x) is 1/4 e^{2x}ln(x)+C