Let \( u^2 = 2x + 1 \Rightarrow u = \sqrt{2x+1} \).

Then \( 2u\,du = 2\,dx \Rightarrow du = dx/u \) and

\[ 6x = 3(2x+1) - 3 = 3u^2 - 3. \]

When \( x = 0 \), \( u = 1 \); when \( x = 1 \), \( u = \sqrt{3} \).

So

\[ \int_0^1 \frac{6x}{\sqrt{2x+1}}dx = \int_1^{\sqrt{3}} (3u^2 - 3)\,du = \bigl[u^3 - 3u\bigr]_1^{\sqrt{3}} = (3\sqrt{3} - 3\sqrt{3}) - (1 - 3) = 2. \]

Answer (3 s.f.): 2.00.

Let \( u = 1 + x^3 \Rightarrow du = 3x^2 dx \), so \( x^2dx = \tfrac{1}{3}du \).

When \( x = 0 \), \( u = 1 \); when \( x = 2 \), \( u = 9 \).

\[ \int_0^2 x^2\sqrt{1+x^3}\,dx = \int_1^9 \tfrac{1}{3}\sqrt{u}\,du = \tfrac{1}{3}\int_1^9 u^{1/2}du = \tfrac{1}{3}\Bigl[\tfrac{2}{3}u^{3/2}\Bigr]_1^9 = \tfrac{2}{9}(27 - 1) = \tfrac{52}{9}. \]

\( \tfrac{52}{9} \approx 5.78 \) (3 s.f.).

Answer: 5.78.

Use the trig substitution \( x = 3\sin\theta \).

Then \( dx = 3\cos\theta\,d\theta \) and

\[ \sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta. \]

When \( x = 0 \), \( \theta = 0 \); when \( x = \tfrac{3}{2} \), \( \sin\theta = \tfrac{1}{2} \Rightarrow \theta = \tfrac{\pi}{6}. \)

So

\[ \int_0^{3/2} \frac{1}{\sqrt{9 - x^2}}\,dx = \int_0^{\pi/6} d\theta = \Bigl[\theta\Bigr]_0^{\pi/6} = \frac{\pi}{6}. \]

\( \dfrac{\pi}{6} \approx 0.524 \) (3 s.f.).

Answer: 0.524.

Let \( u = 1 + 2\sin x \Rightarrow du = 2\cos x\,dx \), so \( \cos x\,dx = \tfrac{1}{2}du \).

When \( x = 0 \), \( u = 1 \); when \( x = \tfrac{\pi}{6} \), \( u = 1 + 2\cdot\tfrac{1}{2} = 2 \).

\[ \int_0^{\pi/6} \frac{\cos x}{\sqrt{1+2\sin x}}dx = \int_1^2 \tfrac{1}{2}u^{-1/2}du = \tfrac{1}{2}\Bigl[2u^{1/2}\Bigr]_1^2 = [\sqrt{u}]_1^2 = \sqrt{2} - 1. \]

\( \sqrt{2} - 1 \approx 0.414 \) (3 s.f.).

Answer: 0.414.

Note that \( 2x + x^2 = x^2 + 2x \) and \( \dfrac{d}{dx}(x^2 + 2x) = 2x + 2 = 2(1+x) \).

Let \( u = x^2 + 2x \Rightarrow du = 2(1+x)dx \), so \( (1+x)dx = \tfrac{1}{2}du \).

When \( x = 0 \), \( u = 0 \); when \( x = 1 \), \( u = 3 \).

\[ \int_0^1 (1+x)\sqrt{2x+x^2}\,dx = \int_0^3 \tfrac{1}{2}\sqrt{u}\,du = \tfrac{1}{2}\int_0^3 u^{1/2}du = \tfrac{1}{2}\Bigl[\tfrac{2}{3}u^{3/2}\Bigr]_0^3 = \tfrac{1}{3}3^{3/2} = \sqrt{3}. \]

\( \sqrt{3} \approx 1.73 \) (3 s.f.).

Answer: 1.73.