What is the integral of (8x+7)ln(x) ?

The integral of (8x+7)ln(x) is 4x^2ln(x)-2x^2+7xln(x)-7x+C

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integral of (8x+7)ln(x)

\int (8 x + 7) ln (x) d x

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

Solution steps

\int (8 x + 7) ln (x) d x

Apply Integration By Parts

: (8 x + 7) (x ln (x) - x) - \int 8 (x ln (x) - x) d x

= (8 x + 7) (x ln (x) - x) - \int 8 (x ln (x) - x) d x

\int 8 (x ln (x) - x) d x = 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2})

= (8 x + 7) (x ln (x) - x) - 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2})

Simplify

(8 x + 7) (x ln (x) - x) - 8 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2}) : 4 x^{2} ln (x) - 2 x^{2} + 7 x ln (x) - 7 x

= 4 x^{2} ln (x) - 2 x^{2} + 7 x ln (x) - 7 x

Add a constant to the solution

= 4 x^{2} ln (x) - 2 x^{2} + 7 x ln (x) - 7 x + C

\int_{5}^{10} e^{x} d x

y = \frac{1}{x}, at, x = 4

\frac{d}{d x} (a cos (t))

y = x^{2} + 2 x - 3

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

What is the integral of (8x+7)ln(x) ?
The integral of (8x+7)ln(x) is 4x^2ln(x)-2x^2+7xln(x)-7x+C