Question 9
Determine $\displaystyle\int\frac{x+1}{x^2+2x}\,dx$
Not quite. Check it by differentiating: the derivative of $\ln\left(\tfrac{1}{2x+2}\right)$ is $-\tfrac{1}{x+1}$, not the integrand. The log should hold the denominator, $x^2+2x$, not something built from its derivative $2x+2$.
Not quite. $2x+2$ is the derivative of the denominator. It belongs on top of the fraction, not inside the log. The derivative of $\ln(2x+2)$ is $\tfrac{1}{x+1}$, which is not the integrand.
That’s it. Step through the hints below to see why.
Not quite. The derivative of $2\ln(x^2+2x)$ is $\dfrac{2(2x+2)}{x^2+2x}=\dfrac{4(x+1)}{x^2+2x}$, four times the integrand. The top is half the derivative of the bottom, so you multiply by $\tfrac12$, not by 2.
The graph shows $x>0$, where $x^2+2x>0$. If $F(x)$ is the right antiderivative, its derivative lies exactly on the blue curve.
The derivative curve sits exactly on the blue one, so $F'(x)$ is the integrand. That’s option (C).
Here $F'(x)$ is 4 times the integrand, well above the blue curve. That’s option (D).
Very close, but not equal. $F'(x)$ is times the integrand. Use the button below.
Here $F'(x)$ is twice the integrand. The top of the integrand is only half of $2x+2$, so you need half of $\ln(x^2+2x)$.
With $k=0$, $F(x)$ is a constant, so its derivative is 0 everywhere.
$F'(x)$ is times the integrand. Only $k=\tfrac12$ makes that exactly 1.
This is option (B). On the right of this graph $F'(x)=\tfrac{1}{x+1}$ looks almost the same as the blue curve, but the gap between them is $\dfrac{1}{x(x+1)(x+2)}$, which is never 0.
This is option (A), $\ln\left(\tfrac{1}{2x+2}\right)=-\ln(2x+2)$. Its derivative, $-\tfrac{1}{x+1}$, is below the axis, while the integrand is above it.
No value of $k$ works with $\ln(2x+2)$. $F'(x)$ divided by the integrand is $\dfrac{k\,x(x+2)}{(x+1)^2}$, which changes with $x$, so the two curves can never lie on top of each other.
Before you integrate a fraction, differentiate its denominator.
$\dfrac{d}{dx}\left(x^2+2x\right)=2x+2$$=2(x+1)$
The top, $x+1$, is exactly half of that.
Run the log rule backwards. Since $\dfrac{d}{dx}\ln f(x)=\dfrac{f'(x)}{f(x)}$, you get $\displaystyle\int\frac{f'(x)}{f(x)}\,dx=\ln f(x)+c$ for $f(x)>0$.
$\displaystyle\int\frac{x+1}{x^2+2x}\,dx$$=\displaystyle\frac12\int\frac{2x+2}{x^2+2x}\,dx$
$=\tfrac12\ln\left(x^2+2x\right)+c$
Check by differentiating your answer. You should get back the integrand.
$\dfrac{d}{dx}\Bigl[\tfrac12\ln\left(x^2+2x\right)\Bigr]$$=\dfrac12\cdot\dfrac{2x+2}{x^2+2x}$$=\dfrac{x+1}{x^2+2x}$
Option (D) gives 4 times the integrand, and (A) and (B) put the wrong expression inside the log.
$\displaystyle\int\frac{x+1}{x^2+2x}\,dx$$=\displaystyle\frac12\int\frac{2x+2}{x^2+2x}\,dx$$=\dfrac12\ln\left(x^2+2x\right)+c$
The top is half the derivative of the bottom, so the answer is half the log of the bottom.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it