What is the integral of 4xln(7x) ?

The integral of 4xln(7x) is 2x^2ln(7x)-x^2+C

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integral of 4xln(7x)

\int 4 x ln (7 x) d x

2 x^{2} ln (7 x) - x^{2} + C

Solution steps

\int 4 x ln (7 x) d x

Take the constant out:

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

= 4 \cdot \int x ln (7 x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} ln (7 x) - \int \frac{x}{2} d x

= 4 (\frac{1}{2} x^{2} ln (7 x) - \int \frac{x}{2} d x)

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= 4 (\frac{1}{2} x^{2} ln (7 x) - \frac{x ^{2}}{4})

Simplify

4 (\frac{1}{2} x^{2} ln (7 x) - \frac{x ^{2}}{4}) : 2 x^{2} ln (7 x) - x^{2}

= 2 x^{2} ln (7 x) - x^{2}

Add a constant to the solution

= 2 x^{2} ln (7 x) - x^{2} + C

\frac{\partial}{\partial x} (ln (2 x + y + 1))

y^{^{'}} + 3 y = 8 x, y (0) = 4

f (x) = 1 - 9 x^{2}, at x = 3

n = 0 \sum \infty (- \frac{5}{4})^{n}

\frac{d v}{d t} = 9.8 - 0.6