Choose your level and practice the substitution method.
Let \( u = \) the “inside” function
Then \( du = u’\,dx \)
Rewrite the integral entirely in terms of \(u\), integrate, then substitute back.
Evaluate \(\displaystyle\int 2x(x^{2}+5)^{3}\,dx\)
1st: Choose \(u\)
\(u = x^{2} + 5\)
2nd: Find \(du\)
\(du = 2x\,dx\)
3. Substitute
\(\displaystyle\int (x^{2}+5)^{3}\cdot 2x\,dx = \int u^{3}\,du\)
4. Integrate & back-substitute
\(\dfrac{1}{4}u^{4} + C = \dfrac{1}{4}(x^{2}+5)^{4} + C\)
\(\dfrac{1}{4}(x^{2}+5)^{4} + C\)
Clear inside function — direct substitution
\(\displaystyle\int 3x^{2}(x^{3}+1)^{4}\,dx\)
1st: Choose \(u\)
\(u = x^{3} + 1\)
2nd: Find \(du\)
\(du = 3x^{2}\,dx\)
3. Substitute
\(\displaystyle\int (x^{3}+1)^{4}\cdot 3x^{2}\,dx = \int u^{4}\,du\)
4. Integrate & back-substitute
\(\dfrac{1}{5}u^{5} + C = \dfrac{1}{5}(x^{3}+1)^{5} + C\)
\(\dfrac{1}{5}(x^{3}+1)^{5} + C\)
\(\displaystyle\int \cos x\cdot\sin^{5}x\,dx\)
1st: Choose \(u\)
\(u = \sin x\)
2nd: Find \(du\)
\(du = \cos x\,dx\)
3. Substitute
\(\displaystyle\int \sin^{5}x\cdot\cos x\,dx = \int u^{5}\,du\)
4. Integrate & back-substitute
\(\dfrac{1}{6}u^{6} + C = \dfrac{1}{6}\sin^{6}x + C\)
\(\dfrac{1}{6}\sin^{6}x + C\)
\(\displaystyle\int e^{5x}\,dx\)
1st: Choose \(u\)
\(u = 5x\)
2nd: Find \(du\)
\(du = 5\,dx \quad\Rightarrow\quad dx = \dfrac{1}{5}du\)
3. Substitute
\(\displaystyle\int e^{5x}\,dx = \int e^{u}\cdot\dfrac{1}{5}\,du = \dfrac{1}{5}\int e^{u}\,du\)
4. Integrate & back-substitute
\(\dfrac{1}{5}e^{u} + C = \dfrac{1}{5}e^{5x} + C\)
\(\dfrac{1}{5}e^{5x} + C\)
Slight rewrite or coefficient adjustment needed
\(\displaystyle\int \dfrac{x}{1+x^{2}}\,dx\)
1st: Choose \(u\)
\(u = 1 + x^{2}\)
2nd: Find \(du\)
\(du = 2x\,dx \quad\Rightarrow\quad x\,dx = \dfrac{1}{2}du\)
3. Substitute
\(\displaystyle\int \dfrac{x}{1+x^{2}}\,dx = \dfrac{1}{2}\int \dfrac{1}{u}\,du\)
4. Integrate & back-substitute
\(\dfrac{1}{2}\ln|u| + C = \dfrac{1}{2}\ln|1+x^{2}| + C\)
\(\dfrac{1}{2}\ln(1+x^{2}) + C\)
\(\displaystyle\int x\sqrt{x^{2}+9}\,dx\)
1st: Choose \(u\)
\(u = x^{2} + 9\)
2nd: Find \(du\)
\(du = 2x\,dx \quad\Rightarrow\quad x\,dx = \dfrac{1}{2}du\)
3. Substitute
\(\displaystyle\int x\sqrt{x^{2}+9}\,dx = \dfrac{1}{2}\int \sqrt{u}\,du = \dfrac{1}{2}\int u^{1/2}\,du\)
4. Integrate & back-substitute
\(\dfrac{1}{2}\cdot\dfrac{2}{3}u^{3/2} + C = \dfrac{1}{3}(x^{2}+9)^{3/2} + C\)
\(\dfrac{1}{3}(x^{2}+9)^{3/2} + C\)
\(\displaystyle\int \dfrac{1}{x\ln x}\,dx\)
1st: Choose \(u\)
\(u = \ln x\)
2nd: Find \(du\)
\(du = \dfrac{1}{x}\,dx\)
3. Substitute
\(\displaystyle\int \dfrac{1}{x\ln x}\,dx = \int \dfrac{1}{u}\,du\)
4. Integrate & back-substitute
\(\ln|u| + C = \ln|\ln x| + C\)
\(\ln|\ln x| + C\)
u-substitution HARD problems
Requires algebraic rewrite or careful coefficient handling
\(\displaystyle\int x\sec^{2}(x^{2}-1)\tan^{2}(x^{2}-1)\,dx\)
1st: Let \(w = x^{2}-1\)
\(dw = 2x\,dx \quad\Rightarrow\quad x\,dx = \dfrac{1}{2}dw\)
Rewrite the integral:
\(\displaystyle\int x\sec^{2}(x^{2}-1)\tan^{2}(x^{2}-1)\,dx = \dfrac{1}{2}\int\sec^{2}w\tan^{2}w\,dw\)
2nd: Now let \(u = \tan w\)
\(du = \sec^{2}w\,dw\)
The integral becomes:
\(\dfrac{1}{2}\int u^{2}\,du\)
3. Integrate
\(\dfrac{1}{2}\cdot\dfrac{u^{3}}{3} + C = \dfrac{1}{6}u^{3} + C\)
4. Back-substitute
\(u = \tan w = \tan(x^{2}-1)\)
\(\dfrac{1}{6}\tan^{3}(x^{2}-1) + C\)
Ready to master substitution and the rest of Calculus 2?