Name: Indefinite Integration by Substitution
Uploaded: 2019-07-19
Duration: 25 min 48 s
Description: We know the basic formulae to solve integration, but can we be satisfied with just these many formulae? Don't we need to find out the integration of more fun...

Question

Evaluate the integral $\int_{0}^{2} \frac{4 x + 5}{(2 x + 1) (x + 2)} d x$

Accepted Answer

Let $I = \int_{0}^{2} \frac{4 x + 5}{(2 x + 1) (x + 2)} d x$

Consider, $\frac{4 x + 5}{(2 x + 1) (x + 2)} = \frac{a_{0}}{2 x + 1} + \frac{a_{1}}{x + 2}$

$\Rightarrow 4 x + 5 = a_{0} (x + 2) + a_{1} (2 x + 1)$

$\Rightarrow 4 x + 5 = a_{0} x + 2 a_{0} + 2 a_{1} x + a_{1}$

$\Rightarrow 4 x + 5 = x (a_{0} + 2 a_{1}) + 2 a_{0} + a_{1}$

On comparing the coefficients of like powers of $x$ , we get:

$4 = a_{0} + 2 a_{1}, 5 = 2 a_{0} + a_{1}$

On solving these, we get:

$a_{0} = 2, a_{1} = 1$

Thus, $I = \int_{0}^{2} (\frac{2}{2 x + 1} + \frac{1}{x + 2}) d x$

$\Rightarrow I = {[\frac{2 \ln |2 x + 1|}{2} + \ln |x + 2|]}_{0}^{2}$

$\Rightarrow I = {[\ln |2 x + 1| + \ln |x + 2|]}_{0}^{2}$

$\Rightarrow I = [\ln |2 (2) + 1| + \ln |2 + 2|] - [\ln |2 (0) + 1| + \ln |0 + 2|]$

$\Rightarrow I = [\ln 5 + \ln 4] - [\ln 1 + \ln 2]$

$\Rightarrow I = [\ln (5 \times 4)] - [0 + \ln 2] [∵ \ln (p \times q) = \ln p + \ln q, \ln (1) = 0]$

$\Rightarrow I = \ln 20 - \ln 2$

$\Rightarrow I = \ln (\frac{20}{2}) [∵ \ln (\frac{p}{q}) = \ln p - \ln q]$

$\Rightarrow I = \ln 10$

Evaluate the integral $\int_{0}^{2} \frac{4 x + 5}{(2 x + 1) (x + 2)} d x$

Important Questions on Further Calculus

Learn and Prepare for any exam you want !

Evaluate the integral∫02 4x+5(2x+1)(x+2)dx

Important Questions on Further Calculus

Learn and Prepare for any exam you want !

Evaluate the integral $\int_{0}^{2} \frac{4 x + 5}{(2 x + 1) (x + 2)} d x$