# Integral check. Is partial fractions the only way?

I'm getting a different answer from wolfram and I have no idea where. I have to integrate:

$$\int_0^1 \frac{xdx}{(2x+1)^3}$$

Is partial fractions the only way?

So evaluating the fraction first:

$$\frac{x}{(2x+1)^3} = \frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3}$$

$$x = A(2x+1)^2 + B(2x+1) + C$$

$$x = A(4x^2 + 4x + 1) + 2Bx + B + C$$

$$x = 4Ax^2 + 4Ax + A + 2Bx + B + C$$

$$x = x^2(4A) + x(4A+2B) + A + B + C$$

$$4A = 0$$ and $$4A+2B = 1$$ and $$A + B + C = 0$$ so $$A = 0$$ and $$B=\frac{1}{2}$$ and $$C = \frac{-1}{2}$$

Is the partial fraction part right?

So then I get:

$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{(2(2x+1)^2)} - \int_0^1 \frac{dx}{2(2x+1)^3}$$

for both, using $$u = 2x+1$$ and $$\frac{du}{dx} = 2$$ and $$du = 2dx$$ and $$\frac{du}{2} = dx$$, $$\frac{1}{4} \int \frac{du}{u^2} - \frac{1}{4} \int \frac{du}{u^3}$$

$$= [\frac{1}{4} - u^{-1} - \frac{1}{4} \cdot \frac{1}{-2} u^{-2} ]_1^3$$

I'm going the route of changing the limits to the new u and since $$u = 2x+1$$, when $$x = 0, u = 1$$ and when $$x = 1, u = 3$$. Is this the right path?

finally I get

$$[\frac{-1}{4}u^{-1} + \frac{1}{8}u^{-2} ]_1^3$$

I plug in numbers but I get a different answer than wolfram...

• I can't see any mistake in your work. What's the answer given by wolfram? – Thomas Shelby Feb 21 at 2:49
• They have 1/18 as the answer @ThomasShelby – Jwan622 Feb 21 at 4:48
• Please recheck your calculations. I'm getting $\frac1 {18}$ when I substitute the numbers. – Thomas Shelby Feb 21 at 6:34

$$\int_0^1 \frac{xdx}{(2x+1)^3} = \int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3}$$

## $$\displaystyle \int \dfrac{dx}{2(2x+1)^2}:$$

$$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$$

$$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^2} = \int \dfrac{du}{4u^2} = -\dfrac{1}{4u}$$

$$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$$

$$\qquad \left[-\dfrac{1}{4u} \right]_1^3 = -\dfrac{1}{12} + \dfrac 14 = \dfrac 16$$

## $$\displaystyle \int \dfrac{dx}{2(2x+1)^3}:$$

$$\qquad u=2x+1,\ x=\dfrac{u-1}{2}, \ dx = \dfrac 12 du$$

$$\qquad \displaystyle \int \dfrac{dx}{2(2x+1)^3} = \int \dfrac{du}{4u^3} = -\dfrac{1}{8u^2}$$

$$\qquad x=0 \mapsto u=1, \ x=1 \mapsto u=3$$

$$\qquad \left[-\dfrac{1}{8u^2} \right]_1^3 = -\dfrac{1}{72} + \dfrac 18 = \dfrac 19$$

$$\int_0^1 \frac{dx}{2(2x+1)^2} - \int_0^1 \frac{dx}{2(2x+1)^3} = \dfrac 16 - \dfrac 19 = \dfrac{1}{18}$$

There is also another way to solve for $$A, B$$, and $$C$$.

$$\frac{A}{2x+1} + \frac{B}{(2x+1)^2} + \frac{C}{(2x+1)^3} =\frac{x}{(2x+1)^3}$$

$$A(2x+1)^2 + B(2x+1) + C = x$$

Let $$x = -\dfrac 12$$ and you get \begin{align} C &= -\dfrac 12 \\ A(2x+1)^2 + B(2x+1) - \dfrac 12 &= x \\ 2A(2x+1)^2 + 2B(2x+1) &= 2x+1 \\ 2A(2x+1) + 2B &= 1 \end{align}

Again, let $$x = -\dfrac 12$$ and you get \begin{align} 2B &= 1 \\ B &= \dfrac 12 \\ 2A(2x+1) + 1 &= 1 \\ A &= 0 \end{align}

Hint: Substitute $$x=\tfrac12 u -\tfrac12$$

Addendum: Your original problem was $$\int_0^1\frac{x\; dx}{(2x+1)^3}$$. For this substitution, compute:

Integrand: We have $$x=\tfrac12 u -\tfrac12=\tfrac12(u-1)$$, so $$\frac{x}{(2x+1)^3}=\frac{\tfrac12(u-1)}{(2\cdot\frac12(u-1)+1)^3} =\tfrac12\frac{u-1}{u^3}=\boxed{\tfrac12\left(u^{-2}-u^{-3}\right)}$$

Differential: We have $$x=\tfrac12 u -\tfrac12$$, so $$dx = d\left(\tfrac12 u -\tfrac12\right)=\boxed{\tfrac12du}$$

Limits of integration: The limits for $$x$$ are $$0$$ and $$1$$, so we find the corresponding values of $$u$$: $$x=0\implies 0=\tfrac12 u -\tfrac12\implies u=\boxed{1}$$ $$x=1\implies 1 = \tfrac12 u -\tfrac12 \implies u=\boxed{3}$$ So, replacing the integrand, differential, and limits with the boxed items above we have $$\int_0^1\frac{x\; dx}{(2x+1)^3} = \int_1^3\tfrac12\left(u^{-2}-u^{-3}\right) \tfrac12du$$ $$=\tfrac14\int_1^3\left(u^{-2}-u^{-3}\right) du$$ $$=\left.\frac14\left(-u^{-1}+\tfrac12u^{-2} \right)\right]_1^3$$ $$=\tfrac14[(-\tfrac13+\tfrac1{18})-(-1+\tfrac12)]$$ $$=\tfrac14[-\tfrac{5}{18}+\tfrac12]$$ $$=\tfrac14[\tfrac{4}{18}]$$ $$=\boxed{\tfrac{1}{18}}$$

This is the precise procedure you should follow for any substitution. Don't take shortcuts.

• You beat me to it. – randomgirl Feb 21 at 2:50
• If I go your way, I wind up with $\frac{1}{2} \int \frac{1}{u^2} - \frac{1}{2} \int \frac{1}{u^3}$ which is different than what I have at some point. I can't see my error. – Jwan622 Feb 21 at 5:11
• Falling way short on the "why" and leaning heavily on the "check out this trick"... – MichaelChirico Feb 21 at 9:52
• This is not a "trick" but a recognition of the entire idea behind $u$-substitution: there's a $(2x+1)^3$ in the denominator and $x = (2x+1)'/2$ in the numerator, so it really is a good idea to make the substitution $u = 2x+1$ here. Solving that integral by the method the OP proposed is extremely tedious by comparison. – KCd Feb 21 at 10:37
• It looks like your error is not computing $du$ correctly. – MPW Feb 21 at 14:52

You've computed the integral correctly, probably you made a mistake while substituting in the limits of integration. Everything else upto that point is correct.

Is partial fractions the only way?

No single way is the only way, you have tons of ways to do this integral. Have a look at the following approaches.

1. Manipulation and U-Substitution

• Make the substitution $$\begin{bmatrix}t \\ \mathrm dt \end{bmatrix}=\begin{bmatrix}2x+1 \\ 2\mathrm dx\end{bmatrix}$$
• Use the General Power Rule for Integrals $$I=\int_{0}^{1}\dfrac{x \mathrm dx}{(2x+1)^3}\implies 2I=\int_{0}^{1}\dfrac{2x+1-1}{(2x+1)^3}\mathrm dx$$ $$2I=\int_{0}^{1}\biggl[\dfrac{1}{(2x+1)^2}-\dfrac{1}{(2x+1)^3}\biggr] \mathrm dx$$

$$\implies I=\dfrac{1}{4}\biggl[-\dfrac{1}{t}+\dfrac{1}{2t^2}\biggr]_{1}^{3}=\dfrac{1}{18}$$

1. Trigonometric Substitution

• Make the substitution $$\begin{bmatrix}x \\ \mathrm dx\end{bmatrix}=\begin{bmatrix}1/2\cdot\tan^2 \theta\\\tan\theta\sec^2\theta\mathrm d\theta\end{bmatrix}$$
• Use the Pythagorean Identity involving tangent and secant functions namely $$\sec^2\theta=1+\tan^2\theta$$ to get $$2x+1=\sec^2\theta$$.

$$I=\int_{0}^{1}\dfrac{x\mathrm dx}{(2x+1)^3}=\int_{0}^{\sqrt{2}}\dfrac{\tan^3\theta \sec^2\theta}{\sec^6\theta}\mathrm d\theta$$

Using the definitions of $$\tan\theta=\sin\theta/\cos\theta$$ and $$\sec\theta=1/\cos\theta$$. The integral simplifies to the following form: $$I=\int_{0}^{\sqrt{2}}\sin^3\theta\cos\theta\mathrm d\theta=\int_{0}^{\sqrt{2}}\sin^3\theta \cdot\mathrm d(\sin\theta)$$

Again use the General Power Formula for Integrals to get the following expression:

$$I=\dfrac{1}{8}\biggl[\sin^4\tan^{-1}(\sqrt{2x})\biggr]_{0}^{1}=\dfrac{1}{18}$$

To compute $$\sin\arctan(\sqrt{2})$$, make use of the following easy to prove identity: $$\sin\tan^{-1} x=\dfrac{x}{\sqrt{1+x^2}}$$

A much much easier way to solve it is by using integration by parts.

Hint: Take $$u=x$$, $$dv = \frac{dx}{(2x+1)^3}$$.

• $u$ but then $dv$? – manooooh Feb 21 at 2:49
• @manooooh What do you mean? I don't understand. – Haris Gusic Feb 21 at 2:53
• I am sorry, I understood that you used sub. My apologies. – manooooh Feb 21 at 3:18

1) Observe: $$x= \frac 12 \times \left ( (2x + 1) -1\right).$$

2) Use this to obtain $$\frac{x}{(2x+1)^3} = \frac{1}{2} \times \frac{1}{(2x+1)^2} - \frac 12 \times \frac{1}{(2x+1)^3}.$$

3) Integrate to get

$$\int \frac{x}{(2x+1)^3} dx = -\frac{1}{4} (2x+1)^{-1} + \frac{1}{8} (2x+1)^{-2}.$$

There are many ways to evaluate this integral. One of them is to consider the function $$I(n;a,b)=I=\int_0^1\frac{xdx}{(ax+b)^n}$$ for $$b>0$$, $$0\neq a>-1$$, and $$1,2\neq n>0$$. First we multiply the RHS by $$1=a/a$$: $$I=\frac1a\int_0^1\frac{axdx}{(ax+b)^n}=\frac1a\int_0^1\frac{ax+b-b}{(ax+b)^n}dx$$ $$I=\frac1a\int_0^1\frac{dx}{(ax+b)^{n-1}}-\frac{b}a\int_0^1\frac{dx}{(ax+b)^n}$$ Then we focus on $$J(n;a,b)=J=\int_0^1\frac{dx}{(ax+b)^n}$$ Sub.: $$u=ax+b\Rightarrow \frac{du}a=dx$$ $$x=1\mapsto u=b+a\\ x=0\mapsto u=b$$ thus $$J=\frac1a\int_{b}^{b+a}\frac{du}{u^n}=\frac1{a(1-n)}\left[(b+a)^{1-n}-b^{1-n}\right]$$ And since $$I(n;a,b)=\frac1aJ(n-1;a,b)-\frac{b}aJ(n;a,b)$$ We have that $$I(n;a,b)=\frac1{a^2}\left[\frac{(b+a)^{2-n}-b^{2-n}}{2-n}-b\frac{(b+a)^{1-n}-b^{1-n}}{n-1}\right]$$ And of course your integral is given by $$I(3;2,1)=1/18$$.

No partial fractions needed :)

PS: This even works for fractional values of $$n$$. Check it out here.