What is the integral from 1 to 6 of (x^2+3)/(7x-x^2) ?

The integral from 1 to 6 of (x^2+3)/(7x-x^2) is -5+55/7 ln(6)

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integral from 1 to 6 of (x^2+3)/(7x-x^2)

\int_{1}^{6} \frac{x ^{2} + 3}{7 x - x ^{2}} d x

- 5 + \frac{55}{7} ln (6)

Decimal Notation

9.07811 \dots

Solution steps

\int_{1}^{6} \frac{x ^{2} + 3}{7 x - x ^{2}} d x

Factor

\frac{x ^{2} + 3}{7 x - x ^{2}}

= \int_{1}^{6} \frac{x ^{2} + 3}{- x ( x - 7 )} d x

Take the constant out:

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

= - \int_{1}^{6} \frac{x ^{2} + 3}{x ( x - 7 )} d x

Expand

\frac{x ^{2} + 3}{x ( x - 7 )} : \frac{x ^{2}}{x ( x - 7 )} + \frac{3}{x ( x - 7 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{1}^{6} \frac{x ^{2}}{x ( x - 7 )} d x + \int_{1}^{6} \frac{3}{x ( x - 7 )} d x)

\int_{1}^{6} \frac{x ^{2}}{x ( x - 7 )} d x = 5 - 7 ln (6)

\int_{1}^{6} \frac{3}{x ( x - 7 )} d x = - \frac{6}{7} ln (6)

= - (5 - 7 ln (6) - \frac{6}{7} ln (6))

Simplify

- (5 - 7 ln (6) - \frac{6}{7} ln (6)) : - 5 + \frac{55}{7} ln (6)

= - 5 + \frac{55}{7} ln (6)

\int_{\frac{π}{6}}^{π} sin (x) d x

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

\int \frac{1}{2} + \frac{1}{2} cos (2 x) d x

\int_{1}^{2} (x^{4} + \frac{1}{4 x ^{4}}) d x

x^{4} - 2 x^{4}

What is the integral from 1 to 6 of (x^2+3)/(7x-x^2) ?
The integral from 1 to 6 of (x^2+3)/(7x-x^2) is -5+55/7 ln(6)