What is the integral from 0 to 1 of (30)/(6x^2+7x+1) ?

The integral from 0 to 1 of (30)/(6x^2+7x+1) is -6ln(2/7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (30)/(6x^2+7x+1)

\int_{0}^{1} \frac{30}{6 x ^{2} + 7 x + 1} d x

- 6 ln (\frac{2}{7})

Decimal Notation

7.51657 \dots

Solution steps

\int_{0}^{1} \frac{30}{6 x ^{2} + 7 x + 1} d x

Take the constant out:

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

= 30 \cdot \int_{0}^{1} \frac{1}{6 x ^{2} + 7 x + 1} d x

Complete the square

6 x^{2} + 7 x + 1 : 6 (x + \frac{7}{12})^{2} - \frac{25}{24}

= 30 \cdot \int_{0}^{1} \frac{1}{6 ( x + \frac{7}{12} ) ^{2} - \frac{25}{24}} d x

Apply u-substitution

: \int_{\frac{7}{12}}^{\frac{19}{12}} \frac{24}{144 u ^{2} - 25} d u

= 30 \cdot \int_{\frac{7}{12}}^{\frac{19}{12}} \frac{24}{144 u ^{2} - 25} d u

Take the constant out:

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

= 30 \cdot 24 \cdot \int_{\frac{7}{12}}^{\frac{19}{12}} \frac{1}{144 u ^{2} - 25} d u

Apply Integral Substitution

= 30 \cdot 24 \cdot \int_{\frac{7}{5}}^{\frac{19}{5}} \frac{1}{60 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 30 \cdot 24 \cdot \frac{1}{60} \cdot \int_{\frac{7}{5}}^{\frac{19}{5}} \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 30 \cdot 24 \cdot \frac{1}{60} \cdot \int_{\frac{7}{5}}^{\frac{19}{5}} \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 30 \cdot 24 \cdot \frac{1}{60} (- \int_{\frac{7}{5}}^{\frac{19}{5}} \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 30 \cdot 24 \cdot \frac{1}{60} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{7}{5}}^{\frac{19}{5}})

Simplify

30 \cdot 24 \cdot \frac{1}{60} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{7}{5}}^{\frac{19}{5}}) : - 12 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{7}{5}}^{\frac{19}{5}}

= - 12 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{7}{5}}^{\frac{19}{5}}

Compute the boundaries:

\frac{ln ( \frac{24}{5} ) - ln ( \frac{14}{5} ) - ln ( \frac{12}{5} ) + ln ( \frac{2}{5} )}{2}

= - 12 \cdot \frac{ln ( \frac{24}{5} ) - ln ( \frac{14}{5} ) - ln ( \frac{12}{5} ) + ln ( \frac{2}{5} )}{2}

Simplify

= - 6 ln (\frac{2}{7})

\int (6 x^{2} - 5 x + 3) d x

2 + 6 e^{a x}

\int_{1}^{2} (- 9 x^{2} + 4 x) d x

x \to 0 lim (\frac{4}{x + 4 h})

f (x) = 4 x^{2} + 3 x, (- 3, 27)

What is the integral from 0 to 1 of (30)/(6x^2+7x+1) ?
The integral from 0 to 1 of (30)/(6x^2+7x+1) is -6ln(2/7)