text{જો } 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) ખંડશઃ સંકલનનો ઉપયોગ કરતા $\int u v dx = u \int v dx - \int (u' \int v dx) dx$, જ્યાં $u = (\log x)^3$ અને $v = x^4$: $\int x^4(\log x)^3 dx = \fra

Vedclass · Accepted Answer

$\frac{2}{25}$

Vedclass · Answer

(A) ખંડશઃ સંકલનનો ઉપયોગ કરતા $\int u v dx = u \int v dx - \int (u' \int v dx) dx$, જ્યાં $u = (\log x)^3$ અને $v = x^4$: $\int x^4(\log x)^3 dx = \frac{x^5}{5}(\log x)^3 - \int 3(\log x)^2 \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4(\log x)^2 dx$ ફરીથી ખંડશઃ સંકલન કરતા: $= \frac{x^5}{5}(\log x)^3 - \frac{3}{5} [\frac{x^5}{5}(\log x)^2 - \int 2(\log x) \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} \int x^4 \log x dx$ ફરીથી ખંડશઃ સંકલન કરતા: $= \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} [\frac{x^5}{5} \log x - \int \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{125}x^5 \log x - \frac{6}{125} \int x^4 dx$ $= x^5 [\frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625}] + k$ સહગુણકોની સરખામણી કરતા: $A = \frac{1}{5}, B = -\frac{3}{25}, C = \frac{6}{125}, D = -\frac{6}{625}$ $A + B + C + 5D = \frac{1}{5} - \frac{3}{25} + \frac{6}{125} + 5(-\frac{6}{625}) = \frac{25 - 15 + 6 - 6}{125} = \frac{10}{125} = \frac{2}{25}$

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

Similar Questions

જો $\int x^2 \cos^2 x \, dx = \frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + c$ હોય,તો $f(1) + g(2) + h(\frac{1}{2}) = $

$\int e^{-3 x}\left(x^2+\sin 4 x\right) d x=$

$\int(1+x) \log x \, dx =$

$\int {\frac{{\log x}}{{{{(x + 1)}^2}}}dx} $ ની કિંમત શું છે?

$\int \sin \sqrt{x} \, dx$ નું મૂલ્ય કેટલું થાય?

Vedclass Test Series

Exam Paper Generator

Online Exam Module