# Solving integrals by substitution with exponent $du$

Given the integral $\int_0^4 x^3(x^2 + 1)^{-\frac{1}{2}}dx$, I have tried to choose $u = x^2 + 1, du = 2x\space dx$, thus being left with the integral $\int_1^{17} \frac{1}{2}(du)^ 3\space u^{-\frac{1}{2}}$

Is there a simple way to solve this, or do I need to find a more suitable method?

• If $u = x^2+1$ then $x^2 = u-1$ and $du = 2xdx$. So, you can write $x^3(x^2+1)^{-1/2}$dx as $x^2 (x^2+1)^{-1/2} x dx = \frac{1}{2}(u-1)\sqrt{u}du$.
– Tom
Oct 31 '13 at 12:05
• @Tom: Make it an answer? Oct 31 '13 at 12:06

Let $u = x^2 +1$. Then we have both $x^2 = u-1$ and $\frac{1}{2} du = xdx$. Rewriting the integrand as $x^3(x^2+1)^{-1/2} dx = x^2 (x^2+1)^{-1/2} \,x\,dx = \frac{1}{2}(u-1)(u)^{-1/2}\,du.$ So, your $u$-sub was right on, just needed to take care of that other term in a reasonable way! Your integral should reduce to $$\int_0^4 x^3(x^2+1)^{-1/2} \,dx = \frac{1}{2}\int_1^{17}\frac{u-1}{\sqrt{u}} \,du$$
another way to answer this integral attention when $$u=x^2+1$$, $$u\ge 0$$ note that lower limit is $$0$$ that's not possible because $$0=x^2+1 \rightarrow x=?$$