text{यदि } int x^4(log x)^3 dx = x^5[A(log x) | vedclass

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

Question

(A) खंडशः समाकलन (integration by parts) $\int u v dx = u \int v dx - \int (u' \int v dx) dx$ का उपयोग करते हुए,जहाँ $u = (\log x)^3$ और $v = x^4$: $\i

Vedclass · Accepted Answer

$\frac{2}{25}$

Vedclass · Answer

(A) खंडशः समाकलन (integration by parts) $\int u v dx = u \int v dx - \int (u' \int v dx) dx$ का उपयोग करते हुए,जहाँ $u = (\log x)^3$ और $v = x^4$: $\int x^4(\log x)^3 dx = \frac{x^5}{5}(\log x)^3 - \int 3(\log x)^2 \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4(\log x)^2 dx$ पुनः खंडशः समाकलन करने पर: $= \frac{x^5}{5}(\log x)^3 - \frac{3}{5} [\frac{x^5}{5}(\log x)^2 - \int 2(\log x) \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} \int x^4 \log x dx$ पुनः खंडशः समाकलन करने पर: $= \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} [\frac{x^5}{5} \log x - \int \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{125}x^5 \log x - \frac{6}{125} \int x^4 dx$ $= x^5 [\frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625}] + k$ गुणांकों की तुलना करने पर: $A = \frac{1}{5}, B = -\frac{3}{25}, C = \frac{6}{125}, D = -\frac{6}{625}$ $A + B + C + 5D = \frac{1}{5} - \frac{3}{25} + \frac{6}{125} + 5(-\frac{6}{625}) = \frac{25 - 15 + 6 - 6}{125} = \frac{10}{125} = \frac{2}{25}$

$\text{यदि } \int x^4(\log x)^3 dx = x^5[A(\log x)^3 + B(\log x)^2 + C \log x + D] + k, \text{ तो } A + B + C + 5D = $

Similar Questions

$\int x e^{x} d x$ ज्ञात कीजिए।

यदि $\int x^3 \sin 3x \, dx = \frac{1}{27}[f(x) \cos 3x + g(x) \sin 3x] + c$ है,तो $f(1) + g(1) =$

यदि $\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$ है,तो $f(x)$ का मान ज्ञात कीजिए।

यदि $\int f(x) \, dx = g(x)$ है,तो $\int f^{-1}(x) \, dx$ किसके बराबर है?

$\int e^{2x+3} \sin 6x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module