Integrals Involving Roots — Question 10

PDF ↗

Question 10

Use one substitution to evaluate: ∫x+1x2+2x+5dx.\int\frac{x+1}{\sqrt{x^2+2x+5}}\,dx.

Original worksheet page 1: question and worked solution for 1-5-010
Show solutionHide solution

Question 10 – Solution

Step 1: Substitute the expression inside the square root. u=x2+2x+5.u=x^2+2x+5. Differentiate: du=(2x+2)dx=2(x+1)dx,(x+1)dx=12du.du=(2x+2)\,dx=2(x+1)\,dx, \qquad (x+1)\,dx=\frac12du. Step 2: Rewrite the integral. I=∫x+1x2+2x+5dx=12∫u−1/2du.\begin{align*} I&=\int\frac{x+1}{\sqrt{x^2+2x+5}}\,dx\\ &=\frac12\int u^{-1/2}\,du. \end{align*} Step 3: Integrate and return to xx. I=12(u1/21/2)+C=u1/2+C=x2+2x+5+C.\begin{align*} I&=\frac12\left(\frac{u^{1/2}}{1/2}\right)+C\\ &=u^{1/2}+C\\ &=\sqrt{x^2+2x+5}+C. \end{align*} x2+2x+5+C\boxed{\sqrt{x^2+2x+5}+C}

Original worksheet page 2: question and worked solution for 1-5-010

Original worksheet layout. Use Enlarge or open the PDF for a closer view.