What is the integral of (-19x^2+50x-25)/(x^2(3x-5)) ?

The integral of (-19x^2+50x-25)/(x^2(3x-5)) is 1/3 (-21ln(3|x|)-(15)/x+2ln|3x-5|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-19x^2+50x-25)/(x^2(3x-5))

\int \frac{- 19 x ^{2} + 50 x - 25}{x ^{2} ( 3 x - 5 )} d x

\frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣) + C

Solution steps

\int \frac{- 19 x ^{2} + 50 x - 25}{x ^{2} ( 3 x - 5 )} d x

Apply u-substitution

: \int \frac{- 19 u ^{2} - 40 u + 50}{3 u ( u + 5 ) ^{2}} d u

= \int \frac{- 19 u ^{2} - 40 u + 50}{3 u ( u + 5 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{- 19 u ^{2} - 40 u + 50}{u ( u + 5 ) ^{2}} d u

Expand

\frac{- 19 u ^{2} - 40 u + 50}{u ( u + 5 ) ^{2}} : - \frac{19 u ^{2}}{u ( u + 5 ) ^{2}} - \frac{40 u}{u ( u + 5 ) ^{2}} + \frac{50}{u ( u + 5 ) ^{2}}

Apply the Sum Rule:

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

= \frac{1}{3} (- \int \frac{19 u ^{2}}{u ( u + 5 ) ^{2}} d u - \int \frac{40 u}{u ( u + 5 ) ^{2}} d u + \int \frac{50}{u ( u + 5 ) ^{2}} d u)

\int \frac{19 u ^{2}}{u ( u + 5 ) ^{2}} d u = 19 (ln ∣ u + 5 ∣ + \frac{5}{u + 5})

\int \frac{40 u}{u ( u + 5 ) ^{2}} d u = - \frac{40}{u + 5}

\int \frac{50}{u ( u + 5 ) ^{2}} d u = 50 (\frac{1}{25} ln ∣ u ∣ - \frac{1}{25} ln ∣ u + 5 ∣ + \frac{1}{5 ( u + 5 )})

= \frac{1}{3} (- 19 (ln ∣ u + 5 ∣ + \frac{5}{u + 5}) - (- \frac{40}{u + 5}) + 50 (\frac{1}{25} ln ∣ u ∣ - \frac{1}{25} ln ∣ u + 5 ∣ + \frac{1}{5 ( u + 5 )}))

Substitute back

u = 3 x - 5

= \frac{1}{3} (- 19 (ln ∣ 3 x - 5 + 5 ∣ + \frac{5}{3 x - 5 + 5}) - (- \frac{40}{3 x - 5 + 5}) + 50 (\frac{1}{25} ln ∣ 3 x - 5 ∣ - \frac{1}{25} ln ∣ 3 x - 5 + 5 ∣ + \frac{1}{5 ( 3 x - 5 + 5 )}))

Simplify

\frac{1}{3} (- 19 (ln ∣ 3 x - 5 + 5 ∣ + \frac{5}{3 x - 5 + 5}) - (- \frac{40}{3 x - 5 + 5}) + 50 (\frac{1}{25} ln ∣ 3 x - 5 ∣ - \frac{1}{25} ln ∣ 3 x - 5 + 5 ∣ + \frac{1}{5 ( 3 x - 5 + 5 )})) : \frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣)

= \frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣)

Add a constant to the solution

= \frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣) + C

\int \frac{1}{y ^{2} + 1} d y

\int tan (x) (tan (x) + sec (x)) d x

\int \frac{2 x}{x ^{2} - 9 x} d x

\int - \frac{50}{x ^{2}} d x

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

What is the integral of (-19x^2+50x-25)/(x^2(3x-5)) ?
The integral of (-19x^2+50x-25)/(x^2(3x-5)) is 1/3 (-21ln(3|x|)-(15)/x+2ln|3x-5|)+C