\(u=(x^2+1)^2\)

power rule for antiderivative¶

\(\displaystyle\int x^ndx=\frac{n^{n+1}}{n+1}+C\)

\(\int_{0}^{2}xe^{x^2}dx\)

\(u=x^2\) \(du=2xdx\) \(\frac{1}{2}du=xdx\)

\(=\int_{0}^{2}e^{x^2}xdx\) \(=\int_{x=0}^{x=2}e^u\left(\frac{1}{2}du\right)\) \(=\frac{1}{2}\int_{x=0}^{x=2}e^udu\) \(=\frac{1}{2}e^u\biggr|_{x=0}^{x=2}\) \(=\frac{1}{2}e^{x^2}\biggr|_{0}^{2}\) \(=\frac{1}{2}e^{2^2}-\frac{1}{2}e^0\) \(\frac{1}{2}e^4-\frac{1}{2}\)

OR

\(=\int_{0}^{4}e^u\left(\frac{1}{2}du\right)\) \(=\frac{1}{2}\int_{0}^{4}e^udu\) \(=\frac{1}{2}e^u\biggr|_{0}^{4}\) \(\frac{1}{2}e^4-\frac{1}{2}e^0\) \(\frac{1}{2}e^4-\frac{1}{2}\)

\(u\)-subs for Definite Integrals¶

Method 1:¶

Find antiderivative in terms of original variable, then plug in original bounds of integration

Method 2:¶

After \(u\)-sub, find new bounds in terms of \(u\), then do not convert antiderivatives back to original variable.

Ex: \(\int_{1}^{2}\frac{x}{(x^2+1)^2}dx\)

\(u=x^2+1\) \(du=2xdx\) \(\frac{1}{2}du-xdx\)

\(=\int_{1}^{2}\frac{1}{u^2}(\frac{1}{2}du)\) \(=\frac{1}{2}\int\frac{1}{u^2}du\)

\(=\int_{2}^{5}\frac{1}{u^2}(\frac{1}{2}du)\) \(=\frac{1}{2}\int_{2}^{5}\frac{1}{u^2}du\) \(=\frac{1}{2}\int_{2}^{5}\)

using method 2:

\(u=x^2+1\) \(x=1:u=1^2+1=2\) \(x=2:u=2^2+1=5\)

\(=\int_{x=1}^{x=2}\frac{1}{u^2}\left(\frac{1}{2}du\right)\) \(\frac{1}{2}\int_{x=1}^{x=2}u^{-2}du\)

hokora please finish this later im so tired i need to catch up with the notes thanks

Ex: \(\int\sqrt{1+\sqrt x}dx\)

\(u=1+\sqrt x\) \(u=1+x^{\frac{1}{2}}\) \(du=\frac{1}{2}x^{-\frac{1}{2}}dx\)

what if i solve for \(x\) instead? \(u-1=\sqrt x\) \((u-1)^2=x\) 2(u-1)du=dx

\(=\int\sqrt udx\) \(=\int\sqrt u(2)(u-1)du\) \(=2\int u^{\frac{1}{2}}\left(u^{\frac{2}{2}}-1\right)du\) \(=2\int u^{\frac{3}{2}}-u^{\frac{1}{2}}du\) \(\displaystyle=2\left(\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\) \(\displaystyle=2\left(\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}\right)\) \(\displaystyle=2\left(\frac{2}{5}\left(1+\sqrt{x}\right)^{\frac{5}{2}}-\frac{2}{3}\left(1+\sqrt{x}\right)^{\frac{3}{2}}\right)+C\)

Ex: \(\displaystyle\int(x+7)\sqrt[3]{3-2x}dx\)

\(\int(x+7)\sqrt[3]{x}dx\) \(\int\left(x^1+7\right)x^{\frac{1}{3}}dx\) \(\int x^{\frac{4}{3}}+7x^{\frac{1}{2}}dx\) \(\displaystyle\frac{x^{\frac{7}{3}}}{\frac{7}{3}}+7\frac{x^{\frac{4}{3}}}{\frac{4}{3}}+C\)

\(=\int(x+7)(3-2x)^{\frac{1}{3}}dx\)

\(u=3-2x\) \(du=-2dx\) \(-\frac{1}{2}du=dx\) \(u=3-2x\) \(u-3=-2x\) \(\frac{3-u}{2}=x\)

\(=\int(x+7)u^{\frac{1}{3}}dx\) \(=\int(x+7)u^{\frac{1}{3}}\left(-\frac{1}{2}\right)du\) \(=\int\left(\frac{3-u}{2}+7\right)u^{\frac{1}{3}}\left(-\frac{1}{2}\right)du\) \(=\int\left(\frac{3}{2}-\frac{u}{2}+\frac{14}{2}\right)u^{\frac{1}{3}}\left(-\frac{1}{2}\right)du\) \(=\int\left(\frac{17}{2}-\frac{u}{2}\right)u^{\frac{1}{3}}\left(-\frac{1}{2}\right)du\) \(=-\frac{1}{2}\int\frac{17}{2}u^{\frac{1}{3}}-\frac{u^{\frac{4}{3}}}{7}du\) \(\displaystyle=-\frac{1}{2}\left(\frac{17}{2}*\frac{u^{\frac{1}{3}}}{\frac{4}{3}}-\frac{u^{\frac{4}{3}}}{2*\frac{7}{3}}\right)+C\)

Ex:\(\int_{-1}^{1}\sqrt{x+1}dx\)

\(u=x+1\) \(du=1\)

\(=\int_{-1}^{1}\sqrt udu\) \(=u^{\frac{1}{2}}du\biggr|_{-1}^{1}\) \(\displaystyle=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\biggr|_{x=-1}^{x=1}\) \(\displaystyle=\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}\biggr|_{x=-1}^{x=1}\) \(=\frac{(1+1)^{\frac{3}{2}}}{\frac{3}{2}}-\cancel{\frac{(-1+1)^{\frac{3}{2}}}{\frac{3}{2}}}\) \(=\frac{2^{\frac{3}{2}}}{\frac{3}{2}}\) \(=\frac{2}{3}(2)^{\frac{3}{2}}\)

Ex: \(\int x\sqrt{x+1}dx\)

\(h\)

hokora finish this later please i want to sleep