What is $\int \frac{16 x - 15 y}{32} - 6 d x$ $\int \frac { 16x - 15y } { 32} - 6 dx$ ?

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Answer 1 · 2018-05-04T11:20:06

$\frac{x^{2}}{4} - \frac{15 x y}{32} - 6 x + C$ $x^2/4-(15xy)/32-6x+C$

$\int$ $int_$ $\frac{16 x - 15 y}{32} - 6 d x$ $(16x-15y)/(32)-6 dx$

$\frac{1}{32}$ $1/32$ $\int$ $int_$ $(16 x - 15 y) d x$ $(16x-15y) dx$ $- 6$ $-6$ $\int$ $int_$ $1 d x$ $1 dx$

$\frac{1}{2}$ $1/2$ $\int$ $int_$ $x d x$ $x dx$ $+ (\frac{15 y}{32} - 6)$ $+((15y)/32-6)$ $\int$ $int_$ $1 d x$ $1 dx$

$\frac{x^{2}}{4} + (- \frac{15 y}{32} - 6)$ $x^2/4+(-(15y)/32-6)$ $\int$ $int_$ $1 d x$ $1 dx$

$\frac{x^{2}}{4} + (- \frac{15 y}{32} - 6) x + C$ $x^2/4+(-(15y)/32-6)x+C$

$= \frac{x^{2}}{4} - \frac{15 x y}{32} - 6 x + C$ $=x^2/4-(15xy)/32-6x+C$

What is \int \frac { 16x - 15y } { 32} - 6 dx∫16x−15y32−6dx\int \frac { 16x - 15y } { 32} - 6 dx?