What is the integral of 16xln(7x) ?

The integral of 16xln(7x) is 16(1/2 x^2ln(7x)-(x^2)/4)+C

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integral of 16xln(7x)

\int 16 x ln (7 x) d x

16 (\frac{1}{2} x^{2} ln (7 x) - \frac{x ^{2}}{4}) + C

Solution steps

\int 16 x ln (7 x) d x

Take the constant out:

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

= 16 \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

= 16 (\frac{1}{2} x^{2} ln (7 x) - \int \frac{x}{2} d x)

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

= 16 (\frac{1}{2} x^{2} ln (7 x) - \frac{x ^{2}}{4})

Add a constant to the solution

= 16 (\frac{1}{2} x^{2} ln (7 x) - \frac{x ^{2}}{4}) + C

\int_{0}^{\frac{π}{2}} 11 sin^{2} (2 x) d x

\int \frac{a ^{2} - x ^{2}}{x} d x

x y^{^{'}} + 2 y = 4 x^{2}

\frac{d}{d x} (\frac{- e ^{\frac{1}{x}}}{x ^{2}})

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 x

What is the integral of 16xln(7x) ?
The integral of 16xln(7x) is 16(1/2 x^2ln(7x)-(x^2)/4)+C