Math III Basic indefinite integrals This is where calculus-level integration starts. Learn the basic formulas precisely and get used to rewriting roots and fractions as powers.
Basic, Standard, Advanced: Grade 12 · Term 2
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Guide Examples Practice Related
Write roots and fractions as powers Rewrite x \sqrt{x} x 1 x 2 \frac{1}{x^2} x 2 1 powers , x 1 2 x^{\frac{1}{2}} x 2 1 x − 2 x^{-2} x − 2
∫ x d x = ∫ x 1 2 d x = 2 3 x 3 2 + C = 2 3 x x + C \int \sqrt{x}\,dx = \int x^{\frac{1}{2}}\,dx = \frac{2}{3}x^{\frac{3}{2}} + C = \frac{2}{3}x\sqrt{x} + C ∫ x d x = ∫ x 2 1 d x = 3 2 x 2 3 + C = 3 2 x x + C ∫ d x x 2 = ∫ x − 2 d x = − x − 1 + C = − 1 x + C \int \frac{dx}{x^2} = \int x^{-2}\,dx = -x^{-1} + C = -\frac{1}{x} + C ∫ x 2 d x = ∫ x − 2 d x = − x − 1 + C = − x 1 + C It is usual to write the answer back with roots and fractions, like the question.
Note
The power rule fails only for α = − 1 \alpha = -1 α = − 1 ∫ d x x = log ∣ x ∣ + C \int \frac{dx}{x} = \log|x| + C ∫ x d x = log ∣ x ∣ + C
When the inside is linear For f ( a x + b ) f(ax + b) f ( a x + b ) ∫ f ( a x + b ) d x = 1 a F ( a x + b ) + C \displaystyle\int f(ax+b)\,dx = \frac{1}{a}F(ax+b) + C ∫ f ( a x + b ) d x = a 1 F ( a x + b ) + C F F F f f f ∫ e 2 x d x = 1 2 e 2 x + C \int e^{2x}\,dx = \frac{1}{2}e^{2x} + C ∫ e 2 x d x = 2 1 e 2 x + C ∫ sin 3 x d x = − 1 3 cos 3 x + C \int \sin 3x\,dx = -\frac{1}{3}\cos 3x + C ∫ sin 3 x d x = − 3 1 cos 3 x + C substitution page.
Worked examples
Find the indefinite integral.
∫ 2 x d x \int 2\sqrt{x}\,dx ∫ 2 x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Rewrite roots and fractions as powers x α x^{\alpha} x α
Answer ∫ 2 x d x = 4 3 x x + C \int 2\sqrt{x}\,dx = \frac{4}{3}x\sqrt{x} + C ∫ 2 x d x = 3 4 x x + C (C C C
Solution Write x = x 1 2 \sqrt{x} = x^{\frac{1}{2}} x = x 2 1 ∫ x α d x = x α + 1 α + 1 + C \displaystyle\int x^{\alpha}\,dx = \frac{x^{\alpha+1}}{\alpha+1} + C ∫ x α d x = α + 1 x α + 1 + C α ≠ − 1 \alpha \ne -1 α = − 1
∫ 2 x d x = ∫ 2 x 1 2 d x = 4 3 x 3 2 + C = 4 3 x x + C \begin{aligned}\int 2\sqrt{x}\,dx &= \int 2x^{\frac{1}{2}}\,dx \\ &= \frac{4}{3}x^{\frac{3}{2}} + C \\ &= \frac{4}{3}x\sqrt{x} + C\end{aligned} ∫ 2 x d x = ∫ 2 x 2 1 d x = 3 4 x 2 3 + C = 3 4 x x + C
Find the indefinite integral.
∫ ( x + 2 ) 2 x d x \int \frac{(\sqrt{x} + 2)^{2}}{x}\,dx ∫ x ( x + 2 ) 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Expand the numerator and divide each term by x x x ∫ d x x = 2 x + C \int \frac{dx}{\sqrt{x}} = 2\sqrt{x} + C ∫ x d x = 2 x + C
Answer ∫ ( x + 2 ) 2 x d x = x + 8 x + 4 log x + C \int \frac{(\sqrt{x} + 2)^{2}}{x}\,dx = x + 8\sqrt{x} + 4\log x + C ∫ x ( x + 2 ) 2 d x = x + 8 x + 4 log x + C (C C C
Solution Expanding the numerator and dividing by x x x
( x + 2 ) 2 x = 1 + 4 x + 4 x \begin{aligned}\frac{(\sqrt{x} + 2)^{2}}{x} &= 1 + \frac{4}{\sqrt{x}} + \frac{4}{x}\end{aligned} x ( x + 2 ) 2 = 1 + x 4 + x 4 Using ∫ d x x = 2 x + C \displaystyle\int \frac{dx}{\sqrt{x}} = 2\sqrt{x} + C ∫ x d x = 2 x + C log ∣ x ∣ = log x \log|x| = \log x log ∣ x ∣ = log x x > 0 x > 0 x > 0
∫ ( 1 + 4 x + 4 x ) d x = x + 8 x + 4 log x + C \begin{aligned}\int \left(1 + \frac{4}{\sqrt{x}} + \frac{4}{x}\right)\,dx &= x + 8\sqrt{x} + 4\log x + C\end{aligned} ∫ ( 1 + x 4 + x 4 ) d x = x + 8 x + 4 log x + C
Find the indefinite integral.
∫ e 2 x + 2 e x d x \int \frac{e^{2x} + 2}{e^{x}}\,dx ∫ e x e 2 x + 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Dividing each term by e x e^x e x e x e^x e x e − x e^{-x} e − x
Answer ∫ e 2 x + 2 e x d x = e x − 2 e − x + C \int \frac{e^{2x} + 2}{e^{x}}\,dx = e^{x} - 2e^{-x} + C ∫ e x e 2 x + 2 d x = e x − 2 e − x + C (C C C
Solution Dividing each term of the numerator by e x e^x e x
e 2 x + 2 e x = e x + 2 e − x \begin{aligned}\frac{e^{2x} + 2}{e^{x}} &= e^{x} + 2e^{-x}\end{aligned} e x e 2 x + 2 = e x + 2 e − x Noting that ∫ e − x d x = − e − x + C \displaystyle\int e^{-x}\,dx = -e^{-x} + C ∫ e − x d x = − e − x + C
∫ ( e x + 2 e − x ) d x = e x − 2 e − x + C \begin{aligned}\int (e^{x} + 2e^{-x})\,dx &= e^{x} - 2e^{-x} + C\end{aligned} ∫ ( e x + 2 e − x ) d x = e x − 2 e − x + C Practice problems
Find the indefinite integral.
∫ x 2 3 d x \int \sqrt[3]{x^{2}}\,dx ∫ 3 x 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Rewrite roots and fractions as powers x α x^{\alpha} x α
Answer ∫ x 2 3 d x = 3 5 x x 2 3 + C \int \sqrt[3]{x^{2}}\,dx = \frac{3}{5}x\sqrt[3]{x^{2}} + C ∫ 3 x 2 d x = 5 3 x 3 x 2 + C (C C C
Solution Write x 2 3 = x 2 3 \sqrt[3]{x^{2}} = x^{\frac{2}{3}} 3 x 2 = x 3 2 ∫ x α d x = x α + 1 α + 1 + C \displaystyle\int x^{\alpha}\,dx = \frac{x^{\alpha+1}}{\alpha+1} + C ∫ x α d x = α + 1 x α + 1 + C α ≠ − 1 \alpha \ne -1 α = − 1
∫ x 2 3 d x = ∫ x 2 3 d x = 3 5 x 5 3 + C = 3 5 x x 2 3 + C \begin{aligned}\int \sqrt[3]{x^{2}}\,dx &= \int x^{\frac{2}{3}}\,dx \\ &= \frac{3}{5}x^{\frac{5}{3}} + C \\ &= \frac{3}{5}x\sqrt[3]{x^{2}} + C\end{aligned} ∫ 3 x 2 d x = ∫ x 3 2 d x = 5 3 x 3 5 + C = 5 3 x 3 x 2 + C
Find the indefinite integral.
∫ ( 2 x + 2 x ) 2 d x \int \left(2\sqrt{x} + \frac{2}{\sqrt{x}}\right)^{2}\,dx ∫ ( 2 x + x 2 ) 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint When you expand the square, x ⋅ 1 x = 1 \sqrt{x}\cdot\frac{1}{\sqrt{x}} = 1 x ⋅ x 1 = 1
Answer ∫ ( 2 x + 2 x ) 2 d x = 2 x 2 + 8 x + 4 log x + C \int \left(2\sqrt{x} + \frac{2}{\sqrt{x}}\right)^{2}\,dx = 2x^{2} + 8x + 4\log x + C ∫ ( 2 x + x 2 ) 2 d x = 2 x 2 + 8 x + 4 log x + C (C C C
Solution Expanding,
( 2 x + 2 x ) 2 = 4 x + 8 + 4 x \begin{aligned}\left(2\sqrt{x} + \frac{2}{\sqrt{x}}\right)^{2} &= 4x + 8 + \frac{4}{x}\end{aligned} ( 2 x + x 2 ) 2 = 4 x + 8 + x 4 Since x > 0 x > 0 x > 0 log ∣ x ∣ = log x \log|x| = \log x log ∣ x ∣ = log x
∫ ( 4 x + 8 + 4 x ) d x = 2 x 2 + 8 x + 4 log x + C \begin{aligned}\int \left(4x + 8 + \frac{4}{x}\right)\,dx &= 2x^{2} + 8x + 4\log x + C\end{aligned} ∫ ( 4 x + 8 + x 4 ) d x = 2 x 2 + 8 x + 4 log x + C
Find the indefinite integral.
∫ ( x − 2 ) 3 x 2 d x \int \frac{(x - 2)^{3}}{x^{2}}\,dx ∫ x 2 ( x − 2 ) 3 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Expand the numerator and divide each term by x 2 x^2 x 2
Answer ∫ ( x − 2 ) 3 x 2 d x = 1 2 x 2 − 6 x + 12 log ∣ x ∣ + 8 x + C \int \frac{(x - 2)^{3}}{x^{2}}\,dx = \frac{1}{2}x^{2} - 6x + 12\log|x| + \frac{8}{x} + C ∫ x 2 ( x − 2 ) 3 d x = 2 1 x 2 − 6 x + 12 log ∣ x ∣ + x 8 + C (C C C
Solution Expanding the numerator and dividing by x 2 x^2 x 2
( x − 2 ) 3 x 2 = x − 6 + 12 x − 8 x 2 \begin{aligned}\frac{(x - 2)^{3}}{x^{2}} &= x - 6 + \frac{12}{x} - \frac{8}{x^{2}}\end{aligned} x 2 ( x − 2 ) 3 = x − 6 + x 12 − x 2 8 Using ∫ d x x = log ∣ x ∣ + C \displaystyle\int \frac{dx}{x} = \log|x| + C ∫ x d x = log ∣ x ∣ + C ∫ d x x 2 = − 1 x + C \displaystyle\int \frac{dx}{x^2} = -\frac{1}{x} + C ∫ x 2 d x = − x 1 + C
∫ ( x − 6 + 12 x − 8 x 2 ) d x = 1 2 x 2 − 6 x + 12 log ∣ x ∣ + 8 x + C \begin{aligned}\int \left(x - 6 + \frac{12}{x} - \frac{8}{x^{2}}\right)\,dx &= \frac{1}{2}x^{2} - 6x + 12\log|x| + \frac{8}{x} + C\end{aligned} ∫ ( x − 6 + x 12 − x 2 8 ) d x = 2 1 x 2 − 6 x + 12 log ∣ x ∣ + x 8 + C