If -ln x2 - x dx - xlnx2- x - fx-c- then fx equals-

I = ∫log (x^2 +x)dx Now, integrating by parts taking u=log(x^2+x),v=1 I nbsp;=xlog(x^2+x) ∫(2x+1)1x^2+xxdx =xlog(x^2+x) 2∫2x+2 12x+2dx =xlog(x^2+x) 2∫(1 1 ...

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Standard XII

Mathematics

Integration by Partial Fractions

If ∫ln x2 + x...

Question

If $\int ln (x^{2} + x) d x = x ln (x^{2} + x) + f (x) + c$ , then $f (x)$ equals:

2 x - ln (x + 1)

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- 2 x - ln (x + 1)

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2 x + ln (x + 1)

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- 2 x + ln (x + 1)

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Solution

The correct option is D $- 2 x + ln (x + 1)$
$I = \int log (x^{2} + x) d x$
Now, integrating by parts taking $u = log (x^{2} + x), v = 1$
$I = x log (x^{2} + x) - \int (2 x + 1) \frac{1}{x^{2} + x} x d x = x log (x^{2} + x) - 2 \int \frac{2 x + 2 - 1}{2 x + 2} d x = x log (x^{2} + x) - 2 \int (1 - \frac{1}{2 (x + 1)}) d x = x log (x^{2} + x) - 2 x + log (x + 1) + c$

Equating it to the given integral, we get, $f (x) = - 2 x + log (x + 1)$

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Integration by Partial Fractions

Standard XII Mathematics

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If ∫ln(x2+x)dx=xln(x2+x)+f(x)+c, then f(x) equals:

The correct option is D −2x+ln(x+1)I=∫log(x2+x)dx Now, integrating by parts taking u=log(x2+x),v=1 I=xlog(x2+x)−∫(2x+1)1x2+xxdx=xlog(x2+x)−2∫2x+2−12x+2dx=xlog(x2+x)−2∫(1−12(x+1))dx=xlog(x2+x)−2x+log(x+1)+c Equating it to the given integral, we get, f(x)=−2x+log(x+1)

If $\int ln (x^{2} + x) d x = x ln (x^{2} + x) + f (x) + c$ , then $f (x)$ equals: