Integrate $\int (x+1)^{10} (x+2) \ dx $ by parts

Question

I am trying to compute the following integral by parts.

$$I=\int (x+1)^{10}(x+2) \ dx$$

I put $u=x+2$ and $dv=(x+1)^{10} dx$, then

$$I=\frac{(x+2)(x+1)^{11}}{11}-\frac{1}{11}\frac{(x+1)^{12}}{12}+c$$

Which is obviously not correct. What did I do wrong?

Answer 1

As Thomas Andrews suggested, let $u= x+ 1$. Then $dx= du$ and $x+ 2= u+ 1$ so the integral becomes $$\int u^{10}(u+ 1)du= \int u^{11}+ u^{10} du= \dfrac{u^{12}}{12}+ \dfrac{u^{11}}{11}+ C$$

Substituting back the integral is $$\dfrac{(x+ 1)^{12}}{12}+ \dfrac{(x+ 1)^{11}}{11}+ C$$

Answer 2

Let $t=x+1$. Then we have $$\int(x+1)^{10}(x+2)\,dx=\int t^{10}(t+1)\,dt$$

With the tabular method, we get the following:

\begin{array}{ccc} & \text{D} & \text{I} \\ + & t+1 & t^{10} \\ - & 1 & \frac1{11}t^{11} \\ + & 0 & \frac1{132}t^{12} \end{array}

which results in our integral $$\int t^{10}(t+1)\,dt=\frac1{11}t^{11}(t+1)-\frac1{132}t^{12}+C.$$

With the final substitution, we get $$\frac1{11}t^{11}(t+1)-\frac1{132}t^{12}+C=\frac1{11}(x+1)^{11}(x+2)-\frac1{132}(x+1)^{12}+C$$ as desired.

Answer 3

To ensure that your solution is correct, differentiate your answer and check if the derivate equals the original integrand or not. Hence,

$$\frac{\mathrm dI}{\mathrm dx}= \frac{\mathrm d}{\mathrm dx}\frac{(x+2)(x+1)^{11}}{11}-\frac{1}{11}\frac{\mathrm d}{\mathrm dx}\frac{(x+1)^{12}}{12}+\frac{\mathrm dc}{\mathrm dx} = (x+2)(x+1)^{10} + \frac{(x+1)^{11}}{11}-\frac{(x+1)^{11}}{11} + 0 = (x+1)^{10}(x+2)$$

The derivative of integral matches with the integrand and hence your answer is obviously correct.