Name: Second Fundamental Theorem of Calculus
Uploaded: 2019-07-19
Duration: 5 min 22 s
Description: In this video, we will learn about the evaluating the definite integral by using the second fundamnetal theorem of calculus in details.

Question

Evaluate

\int \frac{3 x - 1}{(x - 1) (x - 2 (x - 3)} d x

Accepted Answer

We have to find $\int \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} d x .$

$\Rightarrow \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \Rightarrow 3 x - 1 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) \Rightarrow 3 x - 1 = A (x^{2} - 5 x + 6) + B (x^{2} - 4 x + 3) + C (x^{2} - 3 x + 2)$

Equating the coefficients of $x$ and constant terms, we get

$\Rightarrow A + B + C = 0, - 5 A - 4 B - 3 C = 3$ and $6 A + 3 B + 2 C = - 1$

$\Rightarrow A = 1, B = - 5, C = 4$

$∴ \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{1}{x - 1} - \frac{5}{x - 2} + \frac{4}{x - 3}$

$\Rightarrow \int \frac{3 x - 1}{(x - 1) (x - 2) (x - 3)} d x = \int \frac{d x}{x - 1} - 5 \int \frac{d x}{x - 2} + 4 \int \frac{d x}{x - 3} = \log (x - 1) - 5 \log (x - 2) + 4 \log (x - 3) + c (∵ \int \frac{d x}{x} = \log (x)) = \log (\frac{(x - 1) {(x - 3)}^{4}}{{(x - 2)}^{5}}) + c (∵ n \log (m) = \log {(m)}^{n}, \log (m) - \log (n) = \log (\frac{m}{n}))$

Evaluate $\int \frac{3 x - 1}{(x - 1) (x - 2 (x - 3)} d x$

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate ∫3x-1(x-1) (x-2 (x-3)dx

Important Questions on Integrals

Learn and Prepare for any exam you want !

Evaluate $\int \frac{3 x - 1}{(x - 1) (x - 2 (x - 3)} d x$