Let f(x)=intleft(frac{2 x^3-3 x^2+4 x-5}{x^2} | vedclass

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

Question

(C) Given: $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$We simplify the integrand: $f(x)=\int\left(2 x-3+\frac{4}{x}-\frac{5}{x^2}\right)

Vedclass · Accepted Answer

$9+4 \log 5$

Vedclass · Answer

(C) Given: $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$We simplify the integrand: $f(x)=\int\left(2 x-3+\frac{4}{x}-\frac{5}{x^2}\right) d x$Integrating term by term: $f(x)=x^2-3 x+4 \log |x|+\frac{5}{x}+C$Given $f(1)=1$, we substitute $x=1$: $1^2-3(1)+4 \log(1)+\frac{5}{1}+C=1$$1-3+0+5+C=1 \Rightarrow 3+C=1 \Rightarrow C=-2$Thus, $f(x)=x^2-3 x+4 \log |x|+\frac{5}{x}-2$Now, calculate $f(5)$: $f(5)=5^2-3(5)+4 \log 5+\frac{5}{5}-2$$f(5)=25-15+4 \log 5+1-2$$f(5)=9+4 \log 5$

Let $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$ and $f(1)=1$. Then $f(5)=$

Similar Questions

If $\int(1-\cos x) \operatorname{cosec}^2 x \, dx = f(x) + c$,then $f(x)$ is equal to

If $\frac{3 \pi}{4} < x < \frac{7 \pi}{4}$, then $\int \left(2^x - \sqrt{1 + \sin 2x} + \frac{1}{x^2} - \frac{1}{x}\right) dx = $

$\int (2\sin x + \frac{1}{x}) \, dx$ is equal to

$\int \frac{f'(x)}{[f(x)]^2} \, dx = $

Integrate the function: $\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module