int frac{x+1}{sqrt{x^2+x+1}} d x= | vedclass

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

Question

(D) ધારો કે $I = \int \frac{x+1}{\sqrt{x^2+x+1}} dx$. અંશને $\frac{1}{2}(2x+1) + \frac{1}{2}$ તરીકે લખી શકાય. તેથી,$I = \int \frac{\frac{1}{2}(2x+1) +

Vedclass · Accepted Answer

$\sqrt{x^2+x+1}+\frac{1}{2} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+c$

Vedclass · Answer

(D) ધારો કે $I = \int \frac{x+1}{\sqrt{x^2+x+1}} dx$. અંશને $\frac{1}{2}(2x+1) + \frac{1}{2}$ તરીકે લખી શકાય. તેથી,$I = \int \frac{\frac{1}{2}(2x+1) + \frac{1}{2}}{\sqrt{x^2+x+1}} dx = \frac{1}{2} \int \frac{2x+1}{\sqrt{x^2+x+1}} dx + \frac{1}{2} \int \frac{1}{\sqrt{x^2+x+1}} dx$. ધારો કે $I_1 = \frac{1}{2} \int \frac{2x+1}{\sqrt{x^2+x+1}} dx$. $u = x^2+x+1$ લેતા,$du = (2x+1)dx$ મળે,તેથી $I_1 = \frac{1}{2} \int u^{-1/2} du = \frac{1}{2} (2\sqrt{u}) = \sqrt{x^2+x+1}$. ધારો કે $I_2 = \frac{1}{2} \int \frac{1}{\sqrt{(x+1/2)^2 + (\sqrt{3}/2)^2}} dx$. સૂત્ર $\int \frac{dx}{\sqrt{x^2+a^2}} = \sinh^{-1}(\frac{x}{a})$ નો ઉપયોગ કરતા,$I_2 = \frac{1}{2} \sinh^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right) = \frac{1}{2} \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)$ મળે. આમ,$I = \sqrt{x^2+x+1} + \frac{1}{2} \sinh^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + C$.

$\int \frac{x+1}{\sqrt{x^2+x+1}} d x=$

Similar Questions

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ ની કિંમત શોધો.

જો $\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx = A\left(\frac{\alpha x-1}{\beta x+3}\right)^B + C,$ જ્યાં $C$ એ સંકલનનો અચળાંક છે,તો $\alpha + \beta + 20AB$ ની કિંમત શોધો.

સંકલન શોધો: $\int \sqrt{x^2+x+1} \, dx$

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

જો $\int \frac{dx}{(x-1)^{3/2}(x-3)^{1/2}} = \sqrt{f(x)} + c$ હોય,તો $f(-1) - f(0) =$ શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module