If int(2x+4)sqrt{x-1}dx = a(x-1)^{5/2} + b(x- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int(2x+4)\sqrt{x-1}dx$. Substitute $t = x-1$,so $x = t+1$ and $dx = dt$. Then $I = \int(2(t+1)+4)\sqrt{t}dt = \int(2t+6)\sqrt{t}dt$. $I

Vedclass · Accepted Answer

$\frac{28}{5}$

Vedclass · Answer

(B) Let $I = \int(2x+4)\sqrt{x-1}dx$. Substitute $t = x-1$,so $x = t+1$ and $dx = dt$. Then $I = \int(2(t+1)+4)\sqrt{t}dt = \int(2t+6)\sqrt{t}dt$. $I = \int(2t^{3/2} + 6t^{1/2})dt$. Integrating term by term: $I = 2 \cdot \frac{t^{5/2}}{5/2} + 6 \cdot \frac{t^{3/2}}{3/2} + c$. $I = \frac{4}{5}t^{5/2} + 4t^{3/2} + c$. Substituting back $t = x-1$: $I = \frac{4}{5}(x-1)^{5/2} + 4(x-1)^{3/2} + c$. Comparing this with $a(x-1)^{5/2} + b(x-1)^{3/2} + c$,we get $a = \frac{4}{5}$ and $b = 4$. Therefore,$2a+b = 2(\frac{4}{5}) + 4 = \frac{8}{5} + 4 = \frac{8+20}{5} = \frac{28}{5}$.

If $\int(2x+4)\sqrt{x-1}dx = a(x-1)^{5/2} + b(x-1)^{3/2} + c$ where $c$ is a constant of integration,then the value of $(2a+b)$ is

Similar Questions

The value of $\int \frac{(x-2)}{\{(x-2)^{2}(x+3)^{7}\}^{1 / 3}} d x$ is

Integrate the function $\sqrt{\sin 2x} \cos 2x$.

If $\int e^x(1+x) \cdot \sec ^2(x e^x) \, dx = f(x) + \text{constant}$,then $f(x)$ is equal to

Integrate the function $\frac{e^{\tan ^{-1} x}}{1+x^{2}}$.

$\int \frac{e^x(1+x)}{\cos ^2(e^x \cdot x)} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module