Evaluate -4 d x15 x2-34 x-15where C is constant of integration

Let I= nbsp;∫4 d x15 x^2 34 x+15 I= ∫ 4d x(5 x 3)(3 x 5) Using partial fraction, we have I= nbsp;∫4 d x(5 x 3)(3 x 5)=A(5x 3)+B(3x 5) ⇒ nbsp;A= 54, B=34 ∴ ...

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Standard VI

Mathematics

Idea of a Set

Evaluate ∫4 d...

Question

Evaluate $\int \frac{4 d x}{15 x^{2} - 34 x + 15}$
(where $C$ is constant of integration)

\frac{1}{4} ln ∣ ∣ ∣ \frac{3 x + 5}{5 x + 3} ∣ ∣ ∣ + C

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\frac{1}{4} ln ∣ ∣ ∣ \frac{5 x + 3}{3 x + 5} ∣ ∣ ∣ + C

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\frac{1}{4} ln ∣ ∣ ∣ \frac{3 x - 5}{5 x - 3} ∣ ∣ ∣ + C

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\frac{1}{4} ln ∣ ∣ ∣ \frac{5 x - 3}{3 x - 5} ∣ ∣ ∣ + C

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Solution

The correct option is C $\frac{1}{4} ln ∣ ∣ ∣ \frac{3 x - 5}{5 x - 3} ∣ ∣ ∣ + C$
Let $I = \int \frac{4 d x}{15 x^{2} - 34 x + 15}$
$I = \int \frac{4 d x}{(5 x - 3) (3 x - 5)}$
Using partial fraction, we have
$I = \int \frac{4 d x}{(5 x - 3) (3 x - 5)} = \frac{A}{(5 x - 3)} + \frac{B}{(3 x - 5)}$
$\Rightarrow A = - \frac{5}{4}, B = \frac{3}{4}$
$∴ I = - \frac{5}{4} \cdot \frac{ln | 5 x - 3 |}{5} + \frac{3}{4} \cdot \frac{ln | 3 x - 5 |}{3} + C$
$\Rightarrow I = \frac{1}{4} ln ∣ ∣ ∣ \frac{3 x - 5}{5 x - 3} ∣ ∣ ∣ + C$

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Evaluate ∫4dx15x2−34x+15 (where C is constant of integration)

The correct option is C 14ln∣∣∣3x−55x−3∣∣∣+CLet I=∫4dx15x2−34x+15 I=∫4dx(5x−3)(3x−5) Using partial fraction, we have I=∫4dx(5x−3)(3x−5)=A(5x−3)+B(3x−5) ⇒A=−54,B=34 ∴I=−54⋅ln|5x−3|5+34⋅ln|3x−5|3+C ⇒I=14ln∣∣∣3x−55x−3∣∣∣+C

Evaluate $\int \frac{4 d x}{15 x^{2} - 34 x + 15}$
(where $C$ is constant of integration)