0 votes 0 votes $\int \frac{x^3}{\sqrt{1+x^2}}.dx$ Calculus integration calculus engineering-mathematics definite-integral + – aditi19 asked Oct 24, 2018 edited Oct 24, 2018 by aditi19 aditi19 458 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments aditi19 commented Oct 24, 2018 reply Follow Share nope answer is $\frac{1}{3}(1+x^2)^{3/2}-(1+x^2)^{1/2}+C$ 0 votes 0 votes Shubhanshu commented Oct 24, 2018 reply Follow Share I am getting $x^2 \sqrt{1+x^2} -\frac{2}{3}\sqrt[3]{1+x^2}$ 0 votes 0 votes adarsh_1997 commented Oct 24, 2018 i edited by adarsh_1997 Oct 24, 2018 reply Follow Share aditi put t=tan(inverse)x in my answer.you will get the same answer use sec(tan−1(x))=√(x^2+1) 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes use sec(tan−1(x))=√x^2+1 prove of this inverse formula is here:https://socratic.org/questions/how-do-you-simplify-sec-tan-1-x adarsh_1997 answered Oct 24, 2018 adarsh_1997 comment Share Follow See all 0 reply Please log in or register to add a comment.