What is the integral from 0 to 1 of 7/((3x+4)(5x+2)) ?

The integral from 0 to 1 of 7/((3x+4)(5x+2)) is 1/2 ln(2)

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integral from 0 to 1 of 7/((3x+4)(5x+2))

\int_{0}^{1} \frac{7}{( 3 x + 4 ) ( 5 x + 2 )} d x

\frac{1}{2} ln (2)

Decimal Notation

0.34657 \dots

Solution steps

\int_{0}^{1} \frac{7}{( 3 x + 4 ) ( 5 x + 2 )} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} \frac{1}{( 3 x + 4 ) ( 5 x + 2 )} d x

Take the partial fraction of

\frac{1}{( 3 x + 4 ) ( 5 x + 2 )} : - \frac{3}{14 ( 3 x + 4 )} + \frac{5}{14 ( 5 x + 2 )}

= 7 \cdot \int_{0}^{1} - \frac{3}{14 ( 3 x + 4 )} + \frac{5}{14 ( 5 x + 2 )} d x

Apply the Sum Rule:

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

= 7 (- \int_{0}^{1} \frac{3}{14 ( 3 x + 4 )} d x + \int_{0}^{1} \frac{5}{14 ( 5 x + 2 )} d x)

\int_{0}^{1} \frac{3}{14 ( 3 x + 4 )} d x = \frac{1}{14} (ln (7) - 2 ln (2))

\int_{0}^{1} \frac{5}{14 ( 5 x + 2 )} d x = \frac{1}{14} (ln (7) - ln (2))

= 7 (- \frac{1}{14} (ln (7) - 2 ln (2)) + \frac{1}{14} (ln (7) - ln (2)))

Simplify

7 (- \frac{1}{14} (ln (7) - 2 ln (2)) + \frac{1}{14} (ln (7) - ln (2))) : \frac{1}{2} ln (2)

= \frac{1}{2} ln (2)

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What is the integral from 0 to 1 of 7/((3x+4)(5x+2)) ?
The integral from 0 to 1 of 7/((3x+4)(5x+2)) is 1/2 ln(2)