What is the integral from 1 to 4 of 1/(4x^2-17x+4) ?

The integral from 1 to 4 of 1/(4x^2-17x+4) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of 1/(4x^2-17x+4)

\int_{1}^{4} \frac{1}{4 x ^{2} - 17 x + 4} d x

diverges

Solution steps

\int_{1}^{4} \frac{1}{4 x ^{2} - 17 x + 4} d x

Complete the square

4 x^{2} - 17 x + 4 : 4 (x - \frac{17}{8})^{2} - \frac{225}{16}

= \int_{1}^{4} \frac{1}{4 ( x - \frac{17}{8} ) ^{2} - \frac{225}{16}} d x

Apply u-substitution

: \int_{- \frac{9}{8}}^{\frac{15}{8}} \frac{16}{64 u ^{2} - 225} d u

= \int_{- \frac{9}{8}}^{\frac{15}{8}} \frac{16}{64 u ^{2} - 225} d u

Take the constant out:

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

= 16 \cdot \int_{- \frac{9}{8}}^{\frac{15}{8}} \frac{1}{64 u ^{2} - 225} d u

Apply Integral Substitution

= 16 \cdot \int_{- \frac{3}{5}}^{1} \frac{1}{120 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 16 \cdot \frac{1}{120} \cdot \int_{- \frac{3}{5}}^{1} \frac{1}{v ^{2} - 1} d v

Factor

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

= 16 \cdot \frac{1}{120} \cdot \int_{- \frac{3}{5}}^{1} \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

= 16 \cdot \frac{1}{120} (- \int_{- \frac{3}{5}}^{1} \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}

= 16 \cdot \frac{1}{120} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{- \frac{3}{5}}^{1})

Simplify

16 \cdot \frac{1}{120} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{- \frac{3}{5}}^{1}) : - \frac{2}{15} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{- \frac{3}{5}}^{1}

= - \frac{2}{15} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{- \frac{3}{5}}^{1}

Compute the boundaries:

diverges

= diverges

\int_{1}^{4} 2 + 3 x d x

\int_{0}^{\frac{π}{2}} 3 cos^{2} (θ) d θ

\int_{0}^{\frac{π}{8}} x sec^{2} (2 x) d x

\int_{0}^{3} 5 x - 9 d x

\int_{0}^{1} - 2 x 9 - x^{2} d x

What is the integral from 1 to 4 of 1/(4x^2-17x+4) ?
The integral from 1 to 4 of 1/(4x^2-17x+4) is diverges