Question

$\int \frac{cosx+xsinx}{x^2+xcosx} dx =$

• A. $log|\frac{xsinx}{x+cosx}|+c$
• B. $log|\frac{x}{x+cosx}|+c$
• C. $log|cosx + xsinx|+c$
• D. $log|x^2 + xcosx|+c$

Answer 1

Answer: (D)

D

Answer 2

The given integral is $I = \int \frac{cosx+xsinx}{x^2+xcosx} dx$.

We observe the structure of the integrand. It is a fraction. Often, integrals of the form $\int \frac{f'(x)}{f(x)} dx$ result in $log|f(x)|+c$. Let's test this possibility.

Let's consider the denominator as $f(x)$. Let $f(x) = x^2+xcosx$. To find $f'(x)$, we differentiate $x^2$ and $xcosx$ with respect to $x$. $\frac{d}{dx}(x^2) = 2x$. For $xcosx$, we use the product rule $(uv)' = u'v + uv'$, where $u=x$ and $v=cosx$. $\frac{d}{dx}(xcosx) = (1)(cosx) + (x)(-sinx) = cosx - xsinx$. So, $f'(x) = 2x + cosx - xsinx$.

The numerator of the given integral is $cosx+xsinx$. Since $f'(x) = 2x + cosx - xsinx$ is not equal to $cosx+xsinx$, the integral is not directly of the form $\int \frac{f'(x)}{f(x)} dx$ with $f(x)$ being the entire denominator. This rules out option D.

Let's re-examine the numerator $cosx+xsinx$. This expression is the derivative of $xsinx$. Let $g(x) = xsinx$. Then $g'(x) = (1)(sinx) + (x)(cosx) = sinx+xcosx$. The numerator $cosx+xsinx$ is identical to $sinx+xcosx$. So, the numerator is $\frac{d}{dx}(xsinx)$.

Now, let's consider the denominator $x^2+xcosx$. We can factor out $x$ from the denominator: $x^2+xcosx = x(x+cosx)$.

So the integral can be written as: $I = \int \frac{\frac{d}{dx}(xsinx)}{x(x+cosx)} dx$.

This form suggests a substitution or a specific derivative pattern. Consider the derivative of a quotient $\frac{u}{v}$. $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

Let's try to see if the integrand can be written in the form $\frac{d}{dx}\left( \frac{xsinx}{x+cosx} \right)$ or similar. Let $y = \frac{xsinx}{x+cosx}$. Using the quotient rule, with $u=xsinx$ and $v=x+cosx$: $u' = sinx+xcosx$ $v' = 1-sinx$ $\frac{dy}{dx} = \frac{(sinx+xcosx)(x+cosx) - (xsinx)(1-sinx)}{(x+cosx)^2}$ $= \frac{xsinx+x^2cosx+sinxcosx+xcos^2x - xsinx+xsin^2x}{(x+cosx)^2}$ $= \frac{x^2cosx+sinxcosx+x(cos^2x+sin^2x)}{(x+cosx)^2}$ $= \frac{x^2cosx+sinxcosx+x}{(x+cosx)^2}$. This is not the given integrand. So option A is incorrect.

Let's consider option B: $log|\frac{x}{x+cosx}|+c$. This means $f(x) = \frac{x}{x+cosx}$. Let's find $f'(x)$ using the quotient rule, with $u=x$ and $v=x+cosx$: $u' = 1$ $v' = 1-sinx$ $f'(x) = \frac{(1)(x+cosx) - (x)(1-sinx)}{(x+cosx)^2}$ $f'(x) = \frac{x+cosx - x + xsinx}{(x+cosx)^2}$ $f'(x) = \frac{cosx+xsinx}{(x+cosx)^2}$. So, $\int \frac{cosx+xsinx}{(x+cosx)^2} dx = log|\frac{x}{x+cosx}|+c$. Our integral is $\int \frac{cosx+xsinx}{x^2+xcosx} dx = \int \frac{cosx+xsinx}{x(x+cosx)} dx$. The denominators are different: $(x+cosx)^2$ vs $x(x+cosx)$. So, option B is not directly correct.

Let's re-examine the integral. It is a standard form for a product rule derivative in disguise. Consider the derivative of $x \cdot (x+cosx)^{-1}$. No.

Let's consider the expression $x(x+cosx)$. The derivative of $x$ is $1$. The derivative of $x+cosx$ is $1-sinx$. The numerator is $cosx+xsinx$.

Let's check the options again. Option D: $log|x^2 + xcosx|+c$. This means the integrand is $\frac{\frac{d}{dx}(x^2+xcosx)}{x^2+xcosx} = \frac{2x+cosx-xsinx}{x^2+xcosx}$. This is not the given integral.

There must be a simpler way. Let's look at the structure again. Numerator: $cosx+xsinx$. Denominator: $x^2+xcosx$. Let's rewrite the numerator: $xsinx+cosx$. Let's rewrite the denominator: $x(x+cosx)$.

Consider the product $x(x+cosx)$. Let's try to relate the numerator to the derivative of some parts of the denominator. The numerator $xsinx+cosx$ is very similar to the derivative of $xsinx$ which is $sinx+xcosx$. It is also similar to the derivative of $xcosx$ which is $cosx-xsinx$.

Let's try a substitution for the denominator. Let $t = x^2+xcosx$. Then $dt = (2x+cosx-xsinx)dx$. This doesn't match the numerator.

Let's consider the possibility that the problem expects us to notice that the numerator is $d/dx(xsinx)$ and the denominator is $x(x+cosx)$. $I = \int \frac{d(xsinx)}{x(x+cosx)}$. This doesn't simplify directly.

What if the numerator was $x(cosx+xsinx)$? No.

Let's re-examine the options and the question. This is a common type of question where the numerator is the derivative of the denominator. If the numerator was $2x+cosx-xsinx$, then the answer would be $log|x^2+xcosx|+c$. Since the given numerator is $cosx+xsinx$, option D is technically incorrect if the question is taken literally.

However, in multiple choice questions, sometimes there's a subtle trick or a slight variation that makes one option correct. Let's consider the possibility that the question intends for the denominator to be the argument of the logarithm. If $f(x) = x^2+xcosx$, then $f'(x) = 2x+cosx-xsinx$. The numerator is $cosx+xsinx$.

Let's consider the possibility of a typo in the question or options. If the question was $\int \frac{2x+cosx-xsinx}{x^2+xcosx} dx$, then the answer would be $log|x^2+xcosx|+c$.

Let's assume there is no typo and try to find a way to get one of the options. The numerator $cosx+xsinx$ is $d/dx(xsinx)$. The denominator $x^2+xcosx$ is $x(x+cosx)$.

Let's look at the structure of the options again. They are all of the form $log|f(x)|+c$. This implies that the integrand is of the form $f'(x)/f(x)$.

Let's try to derive the integrand from option D. If $y = log|x^2+xcosx|+c$, then $\frac{dy}{dx} = \frac{1}{x^2+xcosx} \cdot \frac{d}{dx}(x^2+xcosx) = \frac{2x+cosx-xsinx}{x^2+xcosx}$. This is not the given integrand $\frac{cosx+xsinx}{x^2+xcosx}$.

There is a discrepancy. Let me assume there is a typo in the question and the numerator should be $2x+cosx-xsinx$. If the question is exactly as written, none of the options seem to be correct by direct differentiation.

Let's consider a possible scenario. If we multiply the numerator and denominator by $x$: $\int \frac{xcosx+x^2sinx}{x^3+x^2cosx} dx$. This doesn't simplify.

Let's try to manipulate the given integrand to match the derivative of one of the options. Consider option D: $log|x^2 + xcosx|+c$. Its derivative is $\frac{2x+cosx-xsinx}{x^2+xcosx}$. The given numerator is $cosx+xsinx$. Can $cosx+xsinx$ be equal to $2x+cosx-xsinx$? $xsinx = 2x-xsinx$ $2xsinx = 2x$ $sinx = 1$. This is not generally true.

This problem seems to have an error in the question or options. However, in competitive exams, sometimes the "closest" option is chosen or there's a very subtle manipulation required.

Let's re-verify the derivative of $x^2+xcosx$. $f(x) = x^2+xcosx$. $f'(x) = 2x + (cosx - xsinx) = 2x+cosx-xsinx$. The numerator is $cosx+xsinx$. These are clearly not the same.

Given the options are all $log|...|+c$, the most common pattern is $f'(x)/f(x)$. If we assume the denominator is $f(x) = x^2+xcosx$, then $f'(x) = 2x+cosx-xsinx$. The numerator provided is $cosx+xsinx$. If the numerator had been $2x+cosx-xsinx$, then option D would be correct. Given the options, option D is the most plausible intended answer, assuming a typo in the numerator. Without assuming a typo, none of the options are correct.

Assuming the question intends for the integral to be of the form $\int \frac{f'(x)}{f(x)} dx$, and $f(x)$ is the denominator, then the numerator should be $2x+cosx-xsinx$. If the question intends for the answer to be $log|x^2+xcosx|+c$, then the original integral should have been $\int \frac{2x+cosx-xsinx}{x^2+xcosx} dx$.

However, if we strictly adhere to the given problem statement: $I = \int \frac{cosx+xsinx}{x^2+xcosx} dx$. Let $u = x^2+xcosx$. Then $du = (2x+cosx-xsinx)dx$. The numerator is $cosx+xsinx$. We can write $cosx+xsinx = (2x+cosx-xsinx) - 2x + 2xsinx$. This doesn't seem to lead to a simple integral.

Given that this is a multiple-choice question, and option D matches the form $log|denominator|+c$, it is highly probable that the question intended for the numerator to be the derivative of the denominator. This is a very common type of integration problem. The difference between the given numerator and the derivative of the denominator is $2x-2xsinx$. It's possible the question setter made an error in the numerator.

If we proceed with the assumption that the question intended for option D to be correct, then the integral should have been: $\int \frac{2x+cosx-xsinx}{x^2+xcosx} dx$. In this case, let $t = x^2+xcosx$. Then $dt = (2x+cosx-xsinx)dx$. The integral becomes $\int \frac{dt}{t} = log|t|+c = log|x^2+xcosx|+c$.

Given the options, and the common patterns in integration questions, it is highly likely that there is a typo in the numerator of the question. If the numerator was $2x+cosx-xsinx$, then option D would be correct.

Final decision: Assuming a typo in the question and that the numerator was intended to be the derivative of the denominator, then option D is the answer. This is a common situation in multiple-choice questions where the most obvious pattern is tested.